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Three Dimensional Geometry

121256 If the line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is parallel to the line $x - 1 = y - 2 = z - 3$ , then $λ + μ =$

1 1

2 4

3 2

4 3

Explanation:

B The equation of line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is
$\frac{x - λ}{λ - 1} = \frac{y - 2}{2 - 2} = \frac{z - 3}{3 - μ}$ i.e. $\frac{x - λ}{λ - 1} = \frac{y - 2}{0} = \frac{z - 3}{3 - μ}$ ...
The line (i) is parallel to the line $x - 1 = y - 2 = z - 3$
$∴ λ - 1 = 0 = 3 - μ \Rightarrow λ = 1$ and $μ = 3$
Hence, $λ + μ = 1 + 3 = 4$

MHT CET-2019

Three Dimensional Geometry

121257 Equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ which are at a distance of one unit form the point $(1, 2, 3)$ are

1

x + 2 y + 2 z = 6, x + 2 y + 2 z = 0

2

x - 2 y + 2 z = 0, x - 2 y + 2 z - 6 = 0

3

x - 2 y - 6 = 0, x - 2 y + z = 6

4

x + 2 y + 2 z = - 6, x + 2 y + 2 z = 5

Explanation:

B The equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ is
$x - 2 y + 2 z + λ = 0$
The required planes are at a distance of one unit from the point $(1, 2, 3)$ .
$| \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} | = 1$
$| \frac{1 (1) + (- 2) (2) + 2 (3) + λ}{\sqrt{1 + 4 + 4}} | = 1$
$| \frac{1 - 4 + 6 + λ}{3} | = 1$
$3 + λ = \pm 3$
$∴ 3 + λ = 3$ or $3 + λ = - 3$
$λ = 0$ or $λ = - 6$
Hence, the equation of planes are $x - 2 y + 2 z = 0$ and $x - 2 y + 2 z - 6 = 0$

MHT CET-2019

Three Dimensional Geometry

121258 If the foot of the perpendicular drawn form the point $(0, 0, 0)$ to the plane is $(4, - 2, - 5)$ , then the equation of the plane is

1

4 x - 2 y + 5 z = - 5

2

4 x - 2 y - 5 z = 45

3

4 x + 2 y - 5 z = 37

4

4 x + 2 y + 5 z = - 13

Explanation:

B The required plane passes through the point $M (4, - 2, - 5)$ and is perpendicular to $\overset{―}{OM}$
$∴ \overset{―}{a} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ and $\overset{―}{OM} = \overset{―}{n} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ , which is normal to the plane.
original image
$∴ \overset{―}{a} = \overset{―}{n}$
$∴$ The equation of the plane is
$\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$
$∴ \overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 k) = (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k})$
$= 16 + 4 + 25$
$\overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45$
$Taking \overset{―}{r} = x \hat{i} + y + j \hat{k}$
$(x \hat{i} + y + j \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45 \Rightarrow 4 x - 2 y - 5 z = 45$ Taking $\overset{―}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

MHT CET-2019

Three Dimensional Geometry

121260 If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $x - 3 = \frac{y - k}{2} = z$ interest, then the value of $k$ is

1

\frac{9}{2}

2

\frac{1}{2}

3

\frac{5}{2}

4

\frac{7}{2}

Explanation:

A
Points on the line are $(1, - 1, 1)$ and $(3, k, 0)$ and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
$∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \Rightarrow | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0$
$∴ 2 (- 5) - (k + 1) (- 2) - 1 (1) = 0 \Rightarrow - 11 + 2 k + 2 = 0 \Rightarrow k = \frac{9}{2}$

MHT CET-2018

Three Dimensional Geometry

121261 If planes $x - c y - b y = 0, c x - y + a z = 0$ and $b x$ $+ a y - z = 0$ pass through a straight line, then $a^{2} + b^{2} + c^{2} =$

1

1 - abc

2 abc -1

3 1-2abc

4

2 abc - 1

Explanation:

C Given,
Planes are -
$x - c y - b z = 0$
$c x - y + a z = 0$
$b x + a y - z = 0$
$∴$ Planes are concurrent
$| \begin{array}{ccc} 1 & - c & - b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$
$1 (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$1 - a^{2} - c^{2} - a b c - a b c - b^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 1 - 2 a b c$

MHT CET-2018

Three Dimensional Geometry

121256 If the line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is parallel to the line $x - 1 = y - 2 = z - 3$ , then $λ + μ =$

1 1

2 4

3 2

4 3

Explanation:

B The equation of line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is
$\frac{x - λ}{λ - 1} = \frac{y - 2}{2 - 2} = \frac{z - 3}{3 - μ}$ i.e. $\frac{x - λ}{λ - 1} = \frac{y - 2}{0} = \frac{z - 3}{3 - μ}$ ...
The line (i) is parallel to the line $x - 1 = y - 2 = z - 3$
$∴ λ - 1 = 0 = 3 - μ \Rightarrow λ = 1$ and $μ = 3$
Hence, $λ + μ = 1 + 3 = 4$

MHT CET-2019

Three Dimensional Geometry

121257 Equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ which are at a distance of one unit form the point $(1, 2, 3)$ are

1

x + 2 y + 2 z = 6, x + 2 y + 2 z = 0

2

x - 2 y + 2 z = 0, x - 2 y + 2 z - 6 = 0

3

x - 2 y - 6 = 0, x - 2 y + z = 6

4

x + 2 y + 2 z = - 6, x + 2 y + 2 z = 5

Explanation:

B The equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ is
$x - 2 y + 2 z + λ = 0$
The required planes are at a distance of one unit from the point $(1, 2, 3)$ .
$| \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} | = 1$
$| \frac{1 (1) + (- 2) (2) + 2 (3) + λ}{\sqrt{1 + 4 + 4}} | = 1$
$| \frac{1 - 4 + 6 + λ}{3} | = 1$
$3 + λ = \pm 3$
$∴ 3 + λ = 3$ or $3 + λ = - 3$
$λ = 0$ or $λ = - 6$
Hence, the equation of planes are $x - 2 y + 2 z = 0$ and $x - 2 y + 2 z - 6 = 0$

MHT CET-2019

Three Dimensional Geometry

121258 If the foot of the perpendicular drawn form the point $(0, 0, 0)$ to the plane is $(4, - 2, - 5)$ , then the equation of the plane is

1

4 x - 2 y + 5 z = - 5

2

4 x - 2 y - 5 z = 45

3

4 x + 2 y - 5 z = 37

4

4 x + 2 y + 5 z = - 13

Explanation:

B The required plane passes through the point $M (4, - 2, - 5)$ and is perpendicular to $\overset{―}{OM}$
$∴ \overset{―}{a} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ and $\overset{―}{OM} = \overset{―}{n} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ , which is normal to the plane.
original image
$∴ \overset{―}{a} = \overset{―}{n}$
$∴$ The equation of the plane is
$\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$
$∴ \overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 k) = (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k})$
$= 16 + 4 + 25$
$\overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45$
$Taking \overset{―}{r} = x \hat{i} + y + j \hat{k}$
$(x \hat{i} + y + j \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45 \Rightarrow 4 x - 2 y - 5 z = 45$ Taking $\overset{―}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

MHT CET-2019

Three Dimensional Geometry

121260 If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $x - 3 = \frac{y - k}{2} = z$ interest, then the value of $k$ is

1

\frac{9}{2}

2

\frac{1}{2}

3

\frac{5}{2}

4

\frac{7}{2}

Explanation:

A
Points on the line are $(1, - 1, 1)$ and $(3, k, 0)$ and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
$∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \Rightarrow | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0$
$∴ 2 (- 5) - (k + 1) (- 2) - 1 (1) = 0 \Rightarrow - 11 + 2 k + 2 = 0 \Rightarrow k = \frac{9}{2}$

MHT CET-2018

Three Dimensional Geometry

121261 If planes $x - c y - b y = 0, c x - y + a z = 0$ and $b x$ $+ a y - z = 0$ pass through a straight line, then $a^{2} + b^{2} + c^{2} =$

1

1 - abc

2 abc -1

3 1-2abc

4

2 abc - 1

Explanation:

C Given,
Planes are -
$x - c y - b z = 0$
$c x - y + a z = 0$
$b x + a y - z = 0$
$∴$ Planes are concurrent
$| \begin{array}{ccc} 1 & - c & - b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$
$1 (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$1 - a^{2} - c^{2} - a b c - a b c - b^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 1 - 2 a b c$

MHT CET-2018

Three Dimensional Geometry

121256 If the line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is parallel to the line $x - 1 = y - 2 = z - 3$ , then $λ + μ =$

1 1

2 4

3 2

4 3

Explanation:

B The equation of line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is
$\frac{x - λ}{λ - 1} = \frac{y - 2}{2 - 2} = \frac{z - 3}{3 - μ}$ i.e. $\frac{x - λ}{λ - 1} = \frac{y - 2}{0} = \frac{z - 3}{3 - μ}$ ...
The line (i) is parallel to the line $x - 1 = y - 2 = z - 3$
$∴ λ - 1 = 0 = 3 - μ \Rightarrow λ = 1$ and $μ = 3$
Hence, $λ + μ = 1 + 3 = 4$

MHT CET-2019

Three Dimensional Geometry

121257 Equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ which are at a distance of one unit form the point $(1, 2, 3)$ are

1

x + 2 y + 2 z = 6, x + 2 y + 2 z = 0

2

x - 2 y + 2 z = 0, x - 2 y + 2 z - 6 = 0

3

x - 2 y - 6 = 0, x - 2 y + z = 6

4

x + 2 y + 2 z = - 6, x + 2 y + 2 z = 5

Explanation:

B The equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ is
$x - 2 y + 2 z + λ = 0$
The required planes are at a distance of one unit from the point $(1, 2, 3)$ .
$| \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} | = 1$
$| \frac{1 (1) + (- 2) (2) + 2 (3) + λ}{\sqrt{1 + 4 + 4}} | = 1$
$| \frac{1 - 4 + 6 + λ}{3} | = 1$
$3 + λ = \pm 3$
$∴ 3 + λ = 3$ or $3 + λ = - 3$
$λ = 0$ or $λ = - 6$
Hence, the equation of planes are $x - 2 y + 2 z = 0$ and $x - 2 y + 2 z - 6 = 0$

MHT CET-2019

Three Dimensional Geometry

121258 If the foot of the perpendicular drawn form the point $(0, 0, 0)$ to the plane is $(4, - 2, - 5)$ , then the equation of the plane is

1

4 x - 2 y + 5 z = - 5

2

4 x - 2 y - 5 z = 45

3

4 x + 2 y - 5 z = 37

4

4 x + 2 y + 5 z = - 13

Explanation:

B The required plane passes through the point $M (4, - 2, - 5)$ and is perpendicular to $\overset{―}{OM}$
$∴ \overset{―}{a} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ and $\overset{―}{OM} = \overset{―}{n} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ , which is normal to the plane.
original image
$∴ \overset{―}{a} = \overset{―}{n}$
$∴$ The equation of the plane is
$\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$
$∴ \overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 k) = (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k})$
$= 16 + 4 + 25$
$\overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45$
$Taking \overset{―}{r} = x \hat{i} + y + j \hat{k}$
$(x \hat{i} + y + j \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45 \Rightarrow 4 x - 2 y - 5 z = 45$ Taking $\overset{―}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

MHT CET-2019

Three Dimensional Geometry

121260 If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $x - 3 = \frac{y - k}{2} = z$ interest, then the value of $k$ is

1

\frac{9}{2}

2

\frac{1}{2}

3

\frac{5}{2}

4

\frac{7}{2}

Explanation:

A
Points on the line are $(1, - 1, 1)$ and $(3, k, 0)$ and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
$∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \Rightarrow | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0$
$∴ 2 (- 5) - (k + 1) (- 2) - 1 (1) = 0 \Rightarrow - 11 + 2 k + 2 = 0 \Rightarrow k = \frac{9}{2}$

MHT CET-2018

Three Dimensional Geometry

121261 If planes $x - c y - b y = 0, c x - y + a z = 0$ and $b x$ $+ a y - z = 0$ pass through a straight line, then $a^{2} + b^{2} + c^{2} =$

1

1 - abc

2 abc -1

3 1-2abc

4

2 abc - 1

Explanation:

C Given,
Planes are -
$x - c y - b z = 0$
$c x - y + a z = 0$
$b x + a y - z = 0$
$∴$ Planes are concurrent
$| \begin{array}{ccc} 1 & - c & - b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$
$1 (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$1 - a^{2} - c^{2} - a b c - a b c - b^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 1 - 2 a b c$

MHT CET-2018

Three Dimensional Geometry

121256 If the line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is parallel to the line $x - 1 = y - 2 = z - 3$ , then $λ + μ =$

1 1

2 4

3 2

4 3

Explanation:

B The equation of line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is
$\frac{x - λ}{λ - 1} = \frac{y - 2}{2 - 2} = \frac{z - 3}{3 - μ}$ i.e. $\frac{x - λ}{λ - 1} = \frac{y - 2}{0} = \frac{z - 3}{3 - μ}$ ...
The line (i) is parallel to the line $x - 1 = y - 2 = z - 3$
$∴ λ - 1 = 0 = 3 - μ \Rightarrow λ = 1$ and $μ = 3$
Hence, $λ + μ = 1 + 3 = 4$

MHT CET-2019

Three Dimensional Geometry

121257 Equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ which are at a distance of one unit form the point $(1, 2, 3)$ are

1

x + 2 y + 2 z = 6, x + 2 y + 2 z = 0

2

x - 2 y + 2 z = 0, x - 2 y + 2 z - 6 = 0

3

x - 2 y - 6 = 0, x - 2 y + z = 6

4

x + 2 y + 2 z = - 6, x + 2 y + 2 z = 5

Explanation:

B The equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ is
$x - 2 y + 2 z + λ = 0$
The required planes are at a distance of one unit from the point $(1, 2, 3)$ .
$| \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} | = 1$
$| \frac{1 (1) + (- 2) (2) + 2 (3) + λ}{\sqrt{1 + 4 + 4}} | = 1$
$| \frac{1 - 4 + 6 + λ}{3} | = 1$
$3 + λ = \pm 3$
$∴ 3 + λ = 3$ or $3 + λ = - 3$
$λ = 0$ or $λ = - 6$
Hence, the equation of planes are $x - 2 y + 2 z = 0$ and $x - 2 y + 2 z - 6 = 0$

MHT CET-2019

Three Dimensional Geometry

121258 If the foot of the perpendicular drawn form the point $(0, 0, 0)$ to the plane is $(4, - 2, - 5)$ , then the equation of the plane is

1

4 x - 2 y + 5 z = - 5

2

4 x - 2 y - 5 z = 45

3

4 x + 2 y - 5 z = 37

4

4 x + 2 y + 5 z = - 13

Explanation:

B The required plane passes through the point $M (4, - 2, - 5)$ and is perpendicular to $\overset{―}{OM}$
$∴ \overset{―}{a} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ and $\overset{―}{OM} = \overset{―}{n} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ , which is normal to the plane.
original image
$∴ \overset{―}{a} = \overset{―}{n}$
$∴$ The equation of the plane is
$\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$
$∴ \overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 k) = (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k})$
$= 16 + 4 + 25$
$\overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45$
$Taking \overset{―}{r} = x \hat{i} + y + j \hat{k}$
$(x \hat{i} + y + j \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45 \Rightarrow 4 x - 2 y - 5 z = 45$ Taking $\overset{―}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

MHT CET-2019

Three Dimensional Geometry

121260 If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $x - 3 = \frac{y - k}{2} = z$ interest, then the value of $k$ is

1

\frac{9}{2}

2

\frac{1}{2}

3

\frac{5}{2}

4

\frac{7}{2}

Explanation:

A
Points on the line are $(1, - 1, 1)$ and $(3, k, 0)$ and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
$∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \Rightarrow | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0$
$∴ 2 (- 5) - (k + 1) (- 2) - 1 (1) = 0 \Rightarrow - 11 + 2 k + 2 = 0 \Rightarrow k = \frac{9}{2}$

MHT CET-2018

Three Dimensional Geometry

121261 If planes $x - c y - b y = 0, c x - y + a z = 0$ and $b x$ $+ a y - z = 0$ pass through a straight line, then $a^{2} + b^{2} + c^{2} =$

1

1 - abc

2 abc -1

3 1-2abc

4

2 abc - 1

Explanation:

C Given,
Planes are -
$x - c y - b z = 0$
$c x - y + a z = 0$
$b x + a y - z = 0$
$∴$ Planes are concurrent
$| \begin{array}{ccc} 1 & - c & - b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$
$1 (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$1 - a^{2} - c^{2} - a b c - a b c - b^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 1 - 2 a b c$

MHT CET-2018

Three Dimensional Geometry

121256 If the line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is parallel to the line $x - 1 = y - 2 = z - 3$ , then $λ + μ =$

1 1

2 4

3 2

4 3

Explanation:

B The equation of line passing through $A (λ, 2, 3)$ and $B (1, 2, μ)$ is
$\frac{x - λ}{λ - 1} = \frac{y - 2}{2 - 2} = \frac{z - 3}{3 - μ}$ i.e. $\frac{x - λ}{λ - 1} = \frac{y - 2}{0} = \frac{z - 3}{3 - μ}$ ...
The line (i) is parallel to the line $x - 1 = y - 2 = z - 3$
$∴ λ - 1 = 0 = 3 - μ \Rightarrow λ = 1$ and $μ = 3$
Hence, $λ + μ = 1 + 3 = 4$

MHT CET-2019

Three Dimensional Geometry

121257 Equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ which are at a distance of one unit form the point $(1, 2, 3)$ are

1

x + 2 y + 2 z = 6, x + 2 y + 2 z = 0

2

x - 2 y + 2 z = 0, x - 2 y + 2 z - 6 = 0

3

x - 2 y - 6 = 0, x - 2 y + z = 6

4

x + 2 y + 2 z = - 6, x + 2 y + 2 z = 5

Explanation:

B The equation of planes parallel to the plane $x - 2 y + 2 z + 4 = 0$ is
$x - 2 y + 2 z + λ = 0$
The required planes are at a distance of one unit from the point $(1, 2, 3)$ .
$| \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} | = 1$
$| \frac{1 (1) + (- 2) (2) + 2 (3) + λ}{\sqrt{1 + 4 + 4}} | = 1$
$| \frac{1 - 4 + 6 + λ}{3} | = 1$
$3 + λ = \pm 3$
$∴ 3 + λ = 3$ or $3 + λ = - 3$
$λ = 0$ or $λ = - 6$
Hence, the equation of planes are $x - 2 y + 2 z = 0$ and $x - 2 y + 2 z - 6 = 0$

MHT CET-2019

Three Dimensional Geometry

121258 If the foot of the perpendicular drawn form the point $(0, 0, 0)$ to the plane is $(4, - 2, - 5)$ , then the equation of the plane is

1

4 x - 2 y + 5 z = - 5

2

4 x - 2 y - 5 z = 45

3

4 x + 2 y - 5 z = 37

4

4 x + 2 y + 5 z = - 13

Explanation:

B The required plane passes through the point $M (4, - 2, - 5)$ and is perpendicular to $\overset{―}{OM}$
$∴ \overset{―}{a} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ and $\overset{―}{OM} = \overset{―}{n} = 4 \hat{i} - 2 \hat{j} - 5 \hat{k}$ , which is normal to the plane.
original image
$∴ \overset{―}{a} = \overset{―}{n}$
$∴$ The equation of the plane is
$\overset{―}{r} \cdot \overset{―}{n} = \overset{―}{a} \cdot \overset{―}{n}$
$∴ \overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 k) = (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k})$
$= 16 + 4 + 25$
$\overset{―}{r} \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45$
$Taking \overset{―}{r} = x \hat{i} + y + j \hat{k}$
$(x \hat{i} + y + j \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} - 5 \hat{k}) = 45 \Rightarrow 4 x - 2 y - 5 z = 45$ Taking $\overset{―}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

MHT CET-2019

Three Dimensional Geometry

121260 If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $x - 3 = \frac{y - k}{2} = z$ interest, then the value of $k$ is

1

\frac{9}{2}

2

\frac{1}{2}

3

\frac{5}{2}

4

\frac{7}{2}

Explanation:

A
Points on the line are $(1, - 1, 1)$ and $(3, k, 0)$ and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
$∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \Rightarrow | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0$
$∴ 2 (- 5) - (k + 1) (- 2) - 1 (1) = 0 \Rightarrow - 11 + 2 k + 2 = 0 \Rightarrow k = \frac{9}{2}$

MHT CET-2018

Three Dimensional Geometry

121261 If planes $x - c y - b y = 0, c x - y + a z = 0$ and $b x$ $+ a y - z = 0$ pass through a straight line, then $a^{2} + b^{2} + c^{2} =$

1

1 - abc

2 abc -1

3 1-2abc

4

2 abc - 1

Explanation:

C Given,
Planes are -
$x - c y - b z = 0$
$c x - y + a z = 0$
$b x + a y - z = 0$
$∴$ Planes are concurrent
$| \begin{array}{ccc} 1 & - c & - b \\ c & - 1 & a \\ b & a & - 1 \end{array} | = 0$
$1 (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$1 - a^{2} - c^{2} - a b c - a b c - b^{2} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 1 - 2 a b c$

MHT CET-2018

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