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Three Dimensional Geometry

121147 If two distinct point $Q, R$ , lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x$ $- 5 y + 2 z = 0$ and $P Q = P R = \sqrt{18}$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $PQR$ is equal to

1

\frac{2}{3} \sqrt{38}

2

\frac{4}{3} \sqrt{38}

3

\frac{8}{3} \sqrt{38}

4

\sqrt{\frac{152}{3}}

Explanation:

B
original image
Equation of line $L$ is $x = y = z$
$\vec{PQ} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(α - 3) + α + 2 + α - 1 = 0$
$\Rightarrow α = \frac{2}{3}$
$∴ T \equiv (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
$PT = \sqrt{{(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2}}$
$= \sqrt{\frac{38}{3}}$ And, $QT = \frac{4}{\sqrt{3}}$
Therefore, Area $= 2 (\frac{1}{2} \times \frac{4}{\sqrt{3}} \times \frac{\sqrt{38}}{\sqrt{3}})$
$= \frac{4 \sqrt{38}}{3}$

JEE Main-28.06.2022

Three Dimensional Geometry

121148 The distance of the point $P (4, 6 - 2)$ from th line passing the point $(- 3, 2, 3)$ and parallel to line with direction ratios $3, 3, - 1$ is equal to:

1

2 \sqrt{3}

2

\sqrt{14}

3 3

4

\sqrt{6}

Explanation:

B Equation of line is
$\frac{x + 3}{3} = \frac{y - 2}{3} = \frac{x - 3}{- 1} = k$
original image
Any point on this line
$M = (3 k - 3, 3 k + 2, 3 - k)$
$D.R of PM line = (3 k - 3 - 4, 3 k + 2 - 6, 3 - k + 2)$
$= (3 k - 7, 3 k - 4, 5 - k)$
$∴ PM$ is perpendicular to line
$∴ 3 (3 k - 7) + 3 (3 k - 4) - 1 (5 - k) = 0$
$\Rightarrow k = 2,$
So, point $M (3, 8, 1)$ -
$Distance, PM = \sqrt{(3 - 4)^{2} + (8 - 6)^{2} + (1 + 2)^{2}}$
$PM = \sqrt{1 + 4 + 9}$
$\Rightarrow PM = \sqrt{14}$

JEE Main-25.01.2023

Three Dimensional Geometry

121149 Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2}$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3}$
A line $L_{3}$ having direction ratios $1, - 1, - 2$ , intersects $L_{1}$ and $L_{2}$ at the point $P$ and $Q$ respectively. The length of the line segment $P Q$ is

1

3 \sqrt{2}

2 4

3

2 \sqrt{6}

4

4 \sqrt{3}

Explanation:

C We have-
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} = λ$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3} = k$
$For line L_{1}, let P = (2 λ + 1, λ + 3, 2 λ + 2)$
$Q = (k + 2, 2 k + 2, 3 k + 3)$
Direction ratio of line $PQ =$
$= (2 λ + 1 - k - 2, λ + 3 - 2 k - 2, 2 λ + 2 - 3 k - 3)$
$= (2 λ - k - 1, λ - 2 k + 1, 2 λ - 3 k - 1)$
$Given, direction ratio of L_{3} : 1, - 1, - 2$
$L_{3}$ and $P Q$ are same line-
So, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
On taking, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1}$
$3 λ = 3 k$
$λ = k$
On taking, $\frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
$2 λ - 4 k + 2 = 2 λ - 3 k - 1$
$k = 3$
From Eq. (i), $λ = k = 3$
Now, $P (7, 6, 8)$ and $Q (5, 8, 12)$
Length of segment-
$P Q = \sqrt{(7 - 5)^{2} + (6 - 8)^{2} + (8 - 12)^{2}}$
$= \sqrt{4 + 4 + 16} = \sqrt{24} = 2 \sqrt{6}$

JEE Main-25.01.2023

Three Dimensional Geometry

121150 If a plane passes through the point $(- 1, k, 0), (2$ , $k, - 1), (1, 1, 2)$ and is parallel to the line
$\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k^{2} + 1}{(k - 1) (k - 2)}$ is

1

\frac{13}{6}

2

\frac{5}{17}

3

\frac{17}{5}

4

\frac{6}{13}

Explanation:

A Line $L : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
or $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$
Point A ( $- 1, k, 0), B (2, k, - 1)$ and $c (1, 1, 2)$
$\vec{CA} = - 2 \hat{i} + (k - 1) \hat{j} - 2 \hat{k}$
$\vec{CA} \times \vec{CB} = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ - 2 k - 1 - 2 \\ 1 k - 1 - 3 \end{array} |$
$\hat{i} (- 3 k + 3 + 2 k - 2) - \hat{j} (6 + 2) + \hat{k} (- 2 k + 2 - k + 1)$ .
$= (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$
The line: $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$ is perpendicular to normal vector.
$∴ 1 . (1 - k) + 1 (- 8) + (- 1) (3 - 3 k) = 0$
$k = 5$
So, $\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{5^{2} + 1}{(5 - 1) (5 - 2)} = \frac{13}{6}$

JEE Main-30.01.2023

Three Dimensional Geometry

121147 If two distinct point $Q, R$ , lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x$ $- 5 y + 2 z = 0$ and $P Q = P R = \sqrt{18}$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $PQR$ is equal to

1

\frac{2}{3} \sqrt{38}

2

\frac{4}{3} \sqrt{38}

3

\frac{8}{3} \sqrt{38}

4

\sqrt{\frac{152}{3}}

Explanation:

B
original image
Equation of line $L$ is $x = y = z$
$\vec{PQ} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(α - 3) + α + 2 + α - 1 = 0$
$\Rightarrow α = \frac{2}{3}$
$∴ T \equiv (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
$PT = \sqrt{{(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2}}$
$= \sqrt{\frac{38}{3}}$ And, $QT = \frac{4}{\sqrt{3}}$
Therefore, Area $= 2 (\frac{1}{2} \times \frac{4}{\sqrt{3}} \times \frac{\sqrt{38}}{\sqrt{3}})$
$= \frac{4 \sqrt{38}}{3}$

JEE Main-28.06.2022

Three Dimensional Geometry

121148 The distance of the point $P (4, 6 - 2)$ from th line passing the point $(- 3, 2, 3)$ and parallel to line with direction ratios $3, 3, - 1$ is equal to:

1

2 \sqrt{3}

2

\sqrt{14}

3 3

4

\sqrt{6}

Explanation:

B Equation of line is
$\frac{x + 3}{3} = \frac{y - 2}{3} = \frac{x - 3}{- 1} = k$
original image
Any point on this line
$M = (3 k - 3, 3 k + 2, 3 - k)$
$D.R of PM line = (3 k - 3 - 4, 3 k + 2 - 6, 3 - k + 2)$
$= (3 k - 7, 3 k - 4, 5 - k)$
$∴ PM$ is perpendicular to line
$∴ 3 (3 k - 7) + 3 (3 k - 4) - 1 (5 - k) = 0$
$\Rightarrow k = 2,$
So, point $M (3, 8, 1)$ -
$Distance, PM = \sqrt{(3 - 4)^{2} + (8 - 6)^{2} + (1 + 2)^{2}}$
$PM = \sqrt{1 + 4 + 9}$
$\Rightarrow PM = \sqrt{14}$

JEE Main-25.01.2023

Three Dimensional Geometry

121149 Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2}$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3}$
A line $L_{3}$ having direction ratios $1, - 1, - 2$ , intersects $L_{1}$ and $L_{2}$ at the point $P$ and $Q$ respectively. The length of the line segment $P Q$ is

1

3 \sqrt{2}

2 4

3

2 \sqrt{6}

4

4 \sqrt{3}

Explanation:

C We have-
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} = λ$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3} = k$
$For line L_{1}, let P = (2 λ + 1, λ + 3, 2 λ + 2)$
$Q = (k + 2, 2 k + 2, 3 k + 3)$
Direction ratio of line $PQ =$
$= (2 λ + 1 - k - 2, λ + 3 - 2 k - 2, 2 λ + 2 - 3 k - 3)$
$= (2 λ - k - 1, λ - 2 k + 1, 2 λ - 3 k - 1)$
$Given, direction ratio of L_{3} : 1, - 1, - 2$
$L_{3}$ and $P Q$ are same line-
So, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
On taking, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1}$
$3 λ = 3 k$
$λ = k$
On taking, $\frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
$2 λ - 4 k + 2 = 2 λ - 3 k - 1$
$k = 3$
From Eq. (i), $λ = k = 3$
Now, $P (7, 6, 8)$ and $Q (5, 8, 12)$
Length of segment-
$P Q = \sqrt{(7 - 5)^{2} + (6 - 8)^{2} + (8 - 12)^{2}}$
$= \sqrt{4 + 4 + 16} = \sqrt{24} = 2 \sqrt{6}$

JEE Main-25.01.2023

Three Dimensional Geometry

121150 If a plane passes through the point $(- 1, k, 0), (2$ , $k, - 1), (1, 1, 2)$ and is parallel to the line
$\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k^{2} + 1}{(k - 1) (k - 2)}$ is

1

\frac{13}{6}

2

\frac{5}{17}

3

\frac{17}{5}

4

\frac{6}{13}

Explanation:

A Line $L : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
or $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$
Point A ( $- 1, k, 0), B (2, k, - 1)$ and $c (1, 1, 2)$
$\vec{CA} = - 2 \hat{i} + (k - 1) \hat{j} - 2 \hat{k}$
$\vec{CA} \times \vec{CB} = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ - 2 k - 1 - 2 \\ 1 k - 1 - 3 \end{array} |$
$\hat{i} (- 3 k + 3 + 2 k - 2) - \hat{j} (6 + 2) + \hat{k} (- 2 k + 2 - k + 1)$ .
$= (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$
The line: $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$ is perpendicular to normal vector.
$∴ 1 . (1 - k) + 1 (- 8) + (- 1) (3 - 3 k) = 0$
$k = 5$
So, $\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{5^{2} + 1}{(5 - 1) (5 - 2)} = \frac{13}{6}$

JEE Main-30.01.2023

Three Dimensional Geometry

121147 If two distinct point $Q, R$ , lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x$ $- 5 y + 2 z = 0$ and $P Q = P R = \sqrt{18}$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $PQR$ is equal to

1

\frac{2}{3} \sqrt{38}

2

\frac{4}{3} \sqrt{38}

3

\frac{8}{3} \sqrt{38}

4

\sqrt{\frac{152}{3}}

Explanation:

B
original image
Equation of line $L$ is $x = y = z$
$\vec{PQ} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(α - 3) + α + 2 + α - 1 = 0$
$\Rightarrow α = \frac{2}{3}$
$∴ T \equiv (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
$PT = \sqrt{{(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2}}$
$= \sqrt{\frac{38}{3}}$ And, $QT = \frac{4}{\sqrt{3}}$
Therefore, Area $= 2 (\frac{1}{2} \times \frac{4}{\sqrt{3}} \times \frac{\sqrt{38}}{\sqrt{3}})$
$= \frac{4 \sqrt{38}}{3}$

JEE Main-28.06.2022

Three Dimensional Geometry

121148 The distance of the point $P (4, 6 - 2)$ from th line passing the point $(- 3, 2, 3)$ and parallel to line with direction ratios $3, 3, - 1$ is equal to:

1

2 \sqrt{3}

2

\sqrt{14}

3 3

4

\sqrt{6}

Explanation:

B Equation of line is
$\frac{x + 3}{3} = \frac{y - 2}{3} = \frac{x - 3}{- 1} = k$
original image
Any point on this line
$M = (3 k - 3, 3 k + 2, 3 - k)$
$D.R of PM line = (3 k - 3 - 4, 3 k + 2 - 6, 3 - k + 2)$
$= (3 k - 7, 3 k - 4, 5 - k)$
$∴ PM$ is perpendicular to line
$∴ 3 (3 k - 7) + 3 (3 k - 4) - 1 (5 - k) = 0$
$\Rightarrow k = 2,$
So, point $M (3, 8, 1)$ -
$Distance, PM = \sqrt{(3 - 4)^{2} + (8 - 6)^{2} + (1 + 2)^{2}}$
$PM = \sqrt{1 + 4 + 9}$
$\Rightarrow PM = \sqrt{14}$

JEE Main-25.01.2023

Three Dimensional Geometry

121149 Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2}$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3}$
A line $L_{3}$ having direction ratios $1, - 1, - 2$ , intersects $L_{1}$ and $L_{2}$ at the point $P$ and $Q$ respectively. The length of the line segment $P Q$ is

1

3 \sqrt{2}

2 4

3

2 \sqrt{6}

4

4 \sqrt{3}

Explanation:

C We have-
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} = λ$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3} = k$
$For line L_{1}, let P = (2 λ + 1, λ + 3, 2 λ + 2)$
$Q = (k + 2, 2 k + 2, 3 k + 3)$
Direction ratio of line $PQ =$
$= (2 λ + 1 - k - 2, λ + 3 - 2 k - 2, 2 λ + 2 - 3 k - 3)$
$= (2 λ - k - 1, λ - 2 k + 1, 2 λ - 3 k - 1)$
$Given, direction ratio of L_{3} : 1, - 1, - 2$
$L_{3}$ and $P Q$ are same line-
So, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
On taking, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1}$
$3 λ = 3 k$
$λ = k$
On taking, $\frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
$2 λ - 4 k + 2 = 2 λ - 3 k - 1$
$k = 3$
From Eq. (i), $λ = k = 3$
Now, $P (7, 6, 8)$ and $Q (5, 8, 12)$
Length of segment-
$P Q = \sqrt{(7 - 5)^{2} + (6 - 8)^{2} + (8 - 12)^{2}}$
$= \sqrt{4 + 4 + 16} = \sqrt{24} = 2 \sqrt{6}$

JEE Main-25.01.2023

Three Dimensional Geometry

121150 If a plane passes through the point $(- 1, k, 0), (2$ , $k, - 1), (1, 1, 2)$ and is parallel to the line
$\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k^{2} + 1}{(k - 1) (k - 2)}$ is

1

\frac{13}{6}

2

\frac{5}{17}

3

\frac{17}{5}

4

\frac{6}{13}

Explanation:

A Line $L : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
or $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$
Point A ( $- 1, k, 0), B (2, k, - 1)$ and $c (1, 1, 2)$
$\vec{CA} = - 2 \hat{i} + (k - 1) \hat{j} - 2 \hat{k}$
$\vec{CA} \times \vec{CB} = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ - 2 k - 1 - 2 \\ 1 k - 1 - 3 \end{array} |$
$\hat{i} (- 3 k + 3 + 2 k - 2) - \hat{j} (6 + 2) + \hat{k} (- 2 k + 2 - k + 1)$ .
$= (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$
The line: $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$ is perpendicular to normal vector.
$∴ 1 . (1 - k) + 1 (- 8) + (- 1) (3 - 3 k) = 0$
$k = 5$
So, $\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{5^{2} + 1}{(5 - 1) (5 - 2)} = \frac{13}{6}$

JEE Main-30.01.2023

Three Dimensional Geometry

121147 If two distinct point $Q, R$ , lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x$ $- 5 y + 2 z = 0$ and $P Q = P R = \sqrt{18}$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $PQR$ is equal to

1

\frac{2}{3} \sqrt{38}

2

\frac{4}{3} \sqrt{38}

3

\frac{8}{3} \sqrt{38}

4

\sqrt{\frac{152}{3}}

Explanation:

B
original image
Equation of line $L$ is $x = y = z$
$\vec{PQ} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$
$(α - 3) + α + 2 + α - 1 = 0$
$\Rightarrow α = \frac{2}{3}$
$∴ T \equiv (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
$PT = \sqrt{{(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2}}$
$= \sqrt{\frac{38}{3}}$ And, $QT = \frac{4}{\sqrt{3}}$
Therefore, Area $= 2 (\frac{1}{2} \times \frac{4}{\sqrt{3}} \times \frac{\sqrt{38}}{\sqrt{3}})$
$= \frac{4 \sqrt{38}}{3}$

JEE Main-28.06.2022

Three Dimensional Geometry

121148 The distance of the point $P (4, 6 - 2)$ from th line passing the point $(- 3, 2, 3)$ and parallel to line with direction ratios $3, 3, - 1$ is equal to:

1

2 \sqrt{3}

2

\sqrt{14}

3 3

4

\sqrt{6}

Explanation:

B Equation of line is
$\frac{x + 3}{3} = \frac{y - 2}{3} = \frac{x - 3}{- 1} = k$
original image
Any point on this line
$M = (3 k - 3, 3 k + 2, 3 - k)$
$D.R of PM line = (3 k - 3 - 4, 3 k + 2 - 6, 3 - k + 2)$
$= (3 k - 7, 3 k - 4, 5 - k)$
$∴ PM$ is perpendicular to line
$∴ 3 (3 k - 7) + 3 (3 k - 4) - 1 (5 - k) = 0$
$\Rightarrow k = 2,$
So, point $M (3, 8, 1)$ -
$Distance, PM = \sqrt{(3 - 4)^{2} + (8 - 6)^{2} + (1 + 2)^{2}}$
$PM = \sqrt{1 + 4 + 9}$
$\Rightarrow PM = \sqrt{14}$

JEE Main-25.01.2023

Three Dimensional Geometry

121149 Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2}$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3}$
A line $L_{3}$ having direction ratios $1, - 1, - 2$ , intersects $L_{1}$ and $L_{2}$ at the point $P$ and $Q$ respectively. The length of the line segment $P Q$ is

1

3 \sqrt{2}

2 4

3

2 \sqrt{6}

4

4 \sqrt{3}

Explanation:

C We have-
$L_{1} : \frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 2}{2} = λ$
$L_{2} : \frac{x - 2}{2} = \frac{y - 2}{2} = \frac{z - 3}{3} = k$
$For line L_{1}, let P = (2 λ + 1, λ + 3, 2 λ + 2)$
$Q = (k + 2, 2 k + 2, 3 k + 3)$
Direction ratio of line $PQ =$
$= (2 λ + 1 - k - 2, λ + 3 - 2 k - 2, 2 λ + 2 - 3 k - 3)$
$= (2 λ - k - 1, λ - 2 k + 1, 2 λ - 3 k - 1)$
$Given, direction ratio of L_{3} : 1, - 1, - 2$
$L_{3}$ and $P Q$ are same line-
So, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
On taking, $\frac{2 λ - k - 1}{1} = \frac{λ - 2 k + 1}{- 1}$
$3 λ = 3 k$
$λ = k$
On taking, $\frac{λ - 2 k + 1}{- 1} = \frac{2 λ - 3 k - 1}{- 2}$
$2 λ - 4 k + 2 = 2 λ - 3 k - 1$
$k = 3$
From Eq. (i), $λ = k = 3$
Now, $P (7, 6, 8)$ and $Q (5, 8, 12)$
Length of segment-
$P Q = \sqrt{(7 - 5)^{2} + (6 - 8)^{2} + (8 - 12)^{2}}$
$= \sqrt{4 + 4 + 16} = \sqrt{24} = 2 \sqrt{6}$

JEE Main-25.01.2023

Three Dimensional Geometry

121150 If a plane passes through the point $(- 1, k, 0), (2$ , $k, - 1), (1, 1, 2)$ and is parallel to the line
$\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k^{2} + 1}{(k - 1) (k - 2)}$ is

1

\frac{13}{6}

2

\frac{5}{17}

3

\frac{17}{5}

4

\frac{6}{13}

Explanation:

A Line $L : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$
or $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$
Point A ( $- 1, k, 0), B (2, k, - 1)$ and $c (1, 1, 2)$
$\vec{CA} = - 2 \hat{i} + (k - 1) \hat{j} - 2 \hat{k}$
$\vec{CA} \times \vec{CB} = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ - 2 k - 1 - 2 \\ 1 k - 1 - 3 \end{array} |$
$\hat{i} (- 3 k + 3 + 2 k - 2) - \hat{j} (6 + 2) + \hat{k} (- 2 k + 2 - k + 1)$ .
$= (1 - k) \hat{i} - 8 \hat{j} + (3 - 3 k) \hat{k}$
The line: $\frac{x - 1}{1} = \frac{y + 1 / 2}{1} = \frac{z + 1}{- 1}$ is perpendicular to normal vector.
$∴ 1 . (1 - k) + 1 (- 8) + (- 1) (3 - 3 k) = 0$
$k = 5$
So, $\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{5^{2} + 1}{(5 - 1) (5 - 2)} = \frac{13}{6}$

JEE Main-30.01.2023

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