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Three Dimensional Geometry

121375 Let $P$ be the point of intersection of the line $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2}$ and the plane $x + y + z = 2$ . If the distance of the point $P$ from the plane $3 x$ $- 4 y + 12 z = 32$ is $q$ , then $q$ and $2 q$ are the roots of the equation

1

x^{2} = 18 x - 72 = 0

2

x^{2} + 18 x + 72 = 0

3

x^{2} - 18 x + 72 = 0

4

x^{2} + 18 x - 72 = 0

Explanation:

C $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = k$
$∴ x = 3 k - 3, y = k - 2, z = 1 - 2 k$
Since, given that,
$P = (3 k - 3, k - 2, 1 - 2 k)$
Given, be the point of intersection of the line and the plane $x + y = z = 2$
$∴ (3 k - 3) + (k - 2) + (1 - 2 k) = 2$
$2 k - 4 = 2$
$k = 3$
Thus, $P = (6, 1, - 5)$
Now, distance point $P$ from the plane
$3 x - 4 y + 12 z = 32$
$= | \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} | = | \frac{- 78}{\sqrt{169}} |$
$q = | \frac{- 78}{13} | = \frac{78}{13} = 6$
Since, $q = 6$
So, $2 q = 12$
We know that,
roots, $α = q = 6$ and $β = 2 q = 12$
Now, quadratic equation-
$x^{2} - (α + β) x + α β = 0$
$\Rightarrow x^{2} - (12 + 6) x + 12 \times 6 = 0$
$\Rightarrow x^{2} - 18 x + 72 = 0$

JEE Main-10.04.2023

Three Dimensional Geometry

121378 Let the line $\frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5}$ intersect the lines $\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1}$ and $\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $A B$ from the plane $2 x - 2 y + z = 14$ is

1 4

2

\frac{10}{3}

3 3

4

\frac{11}{3}

Explanation:

A Given,
$l_{1} := \frac{x}{1} = \frac{y - 6}{- 2} = \frac{z + 8}{5} = λ$
$l_{2} := \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = μ$
$l_{3} := \frac{x + 3}{6} = \frac{y - 3}{- 3} = \frac{z - 6}{1} = γ$
Intersection of $l_{1}$ and $l_{2}$
$\Rightarrow (λ, - 2 λ + 6, 5 λ - 8) & (4 μ + 5, 3 μ + 7, μ - 2)$
$\Rightarrow λ = 1$ and $μ = - 1$
So, point A $(1, 4, - 3)$
Intersection of (i) and (iii)
$(λ, - 2 λ + 6, 5 λ - 8) & (6 γ - 3, - 3 γ + 3, γ + 6)$
$\Rightarrow λ = 3$ and $γ = 1$
So, point $(B) = (2, 2, 2)$
Perpendicular distance from plane $2 x - 2 y + z = 14$ is
$\Rightarrow d = | \frac{2 (2) - 2 (2) + 2 - 14}{\sqrt{4 + 4 + 1}} |$

JEE Main-10.04.2023

Three Dimensional Geometry

121380 A plane $E$ is perpendicular to the two planes $2 x$ $- 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $3 \sqrt{2}$ , then $(PQ)^{2}$ is equal to

1 9

2 12

3 21

4 33

Explanation:

C Given, equation of planes :
$P_{1} : 2 x - 2 y + z = 0$
$P_{2} : x - y + 2 z = 4$
\& Points: P $(1, - 1, 1), Q (a, a, 2)$
Equation of plane $E$ is
$a (x - 1) + b (y + 1) + c (z - 1) = 0$
Also, it is perpendicular to $P_{1}$ and $P_{2}$
$\Rightarrow 2 x - 2 b + c = 0$
And
$a - b + 2 c = 0$
On solving (i) and (ii)
$\frac{a}{- 4 + 1} = \frac{b}{1 - 4} = \frac{c}{0} = λ$
$\Rightarrow a = 3 λ, b = - 3 λ$ and $c = 0$
\& Equation of plane $E$ -
$- 3 λ (x - 1) - 3 λ (y + 1) + 0 (z - 1) = 0$
Now, distance from plane to $Q$ is -
$d = | \frac{1 \times a + 1 \times a + 0 \times 2}{\sqrt{1^{2} + 1^{2} + 0^{2}}} | = 3 \sqrt{2}$
$\Rightarrow \frac{2 a}{\sqrt{2}} = 3 \sqrt{2}$
$\Rightarrow a = 3$
$∴ Q = (3, 3, 2)$
$Now, (PQ)^{2} = ((3 - 1)^{2} + (3 + 1)^{2} + (2 - 1)^{2})$
$\Rightarrow (PQ)^{2} = 21$ ( $∴$ Given)

JEE Main-25.07.2022

Three Dimensional Geometry

121387 The length of perpendicular from point $\hat{i} + 2 \hat{j} + 3 \hat{k}$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

1 6

2 7

3

\sqrt{17}

4

\sqrt{14}

Explanation:

B Given,
Equation of line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ or
$\vec{r} = 6 i + 7 j + 7 k + λ (3 i + 2 j - 2 k)$
Where, $\vec{a} = 6 i + 7 j + 7 k$ and $\vec{b} = 3 i + 2 j - 2 k$
Let $\vec{c} = i + 2 j + 3 k$
$\Rightarrow Required distance = \frac{| (\vec{c} - \vec{a}) \times \vec{b} |}{| \vec{b} |}$
$= | \frac{(- 5 i - 5 j - 4 k) \times (3 i + 2 j - 2 k)}{\sqrt{17}} |$
$= \frac{| 18 i - 22 j + 5 k |}{\sqrt{17}} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{49}$
$\Rightarrow$ Length $(d) = 7$

AMU-2018

Three Dimensional Geometry

121375 Let $P$ be the point of intersection of the line $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2}$ and the plane $x + y + z = 2$ . If the distance of the point $P$ from the plane $3 x$ $- 4 y + 12 z = 32$ is $q$ , then $q$ and $2 q$ are the roots of the equation

1

x^{2} = 18 x - 72 = 0

2

x^{2} + 18 x + 72 = 0

3

x^{2} - 18 x + 72 = 0

4

x^{2} + 18 x - 72 = 0

Explanation:

C $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = k$
$∴ x = 3 k - 3, y = k - 2, z = 1 - 2 k$
Since, given that,
$P = (3 k - 3, k - 2, 1 - 2 k)$
Given, be the point of intersection of the line and the plane $x + y = z = 2$
$∴ (3 k - 3) + (k - 2) + (1 - 2 k) = 2$
$2 k - 4 = 2$
$k = 3$
Thus, $P = (6, 1, - 5)$
Now, distance point $P$ from the plane
$3 x - 4 y + 12 z = 32$
$= | \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} | = | \frac{- 78}{\sqrt{169}} |$
$q = | \frac{- 78}{13} | = \frac{78}{13} = 6$
Since, $q = 6$
So, $2 q = 12$
We know that,
roots, $α = q = 6$ and $β = 2 q = 12$
Now, quadratic equation-
$x^{2} - (α + β) x + α β = 0$
$\Rightarrow x^{2} - (12 + 6) x + 12 \times 6 = 0$
$\Rightarrow x^{2} - 18 x + 72 = 0$

JEE Main-10.04.2023

Three Dimensional Geometry

121378 Let the line $\frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5}$ intersect the lines $\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1}$ and $\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $A B$ from the plane $2 x - 2 y + z = 14$ is

1 4

2

\frac{10}{3}

3 3

4

\frac{11}{3}

Explanation:

A Given,
$l_{1} := \frac{x}{1} = \frac{y - 6}{- 2} = \frac{z + 8}{5} = λ$
$l_{2} := \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = μ$
$l_{3} := \frac{x + 3}{6} = \frac{y - 3}{- 3} = \frac{z - 6}{1} = γ$
Intersection of $l_{1}$ and $l_{2}$
$\Rightarrow (λ, - 2 λ + 6, 5 λ - 8) & (4 μ + 5, 3 μ + 7, μ - 2)$
$\Rightarrow λ = 1$ and $μ = - 1$
So, point A $(1, 4, - 3)$
Intersection of (i) and (iii)
$(λ, - 2 λ + 6, 5 λ - 8) & (6 γ - 3, - 3 γ + 3, γ + 6)$
$\Rightarrow λ = 3$ and $γ = 1$
So, point $(B) = (2, 2, 2)$
Perpendicular distance from plane $2 x - 2 y + z = 14$ is
$\Rightarrow d = | \frac{2 (2) - 2 (2) + 2 - 14}{\sqrt{4 + 4 + 1}} |$

JEE Main-10.04.2023

Three Dimensional Geometry

121380 A plane $E$ is perpendicular to the two planes $2 x$ $- 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $3 \sqrt{2}$ , then $(PQ)^{2}$ is equal to

1 9

2 12

3 21

4 33

Explanation:

C Given, equation of planes :
$P_{1} : 2 x - 2 y + z = 0$
$P_{2} : x - y + 2 z = 4$
\& Points: P $(1, - 1, 1), Q (a, a, 2)$
Equation of plane $E$ is
$a (x - 1) + b (y + 1) + c (z - 1) = 0$
Also, it is perpendicular to $P_{1}$ and $P_{2}$
$\Rightarrow 2 x - 2 b + c = 0$
And
$a - b + 2 c = 0$
On solving (i) and (ii)
$\frac{a}{- 4 + 1} = \frac{b}{1 - 4} = \frac{c}{0} = λ$
$\Rightarrow a = 3 λ, b = - 3 λ$ and $c = 0$
\& Equation of plane $E$ -
$- 3 λ (x - 1) - 3 λ (y + 1) + 0 (z - 1) = 0$
Now, distance from plane to $Q$ is -
$d = | \frac{1 \times a + 1 \times a + 0 \times 2}{\sqrt{1^{2} + 1^{2} + 0^{2}}} | = 3 \sqrt{2}$
$\Rightarrow \frac{2 a}{\sqrt{2}} = 3 \sqrt{2}$
$\Rightarrow a = 3$
$∴ Q = (3, 3, 2)$
$Now, (PQ)^{2} = ((3 - 1)^{2} + (3 + 1)^{2} + (2 - 1)^{2})$
$\Rightarrow (PQ)^{2} = 21$ ( $∴$ Given)

JEE Main-25.07.2022

Three Dimensional Geometry

121387 The length of perpendicular from point $\hat{i} + 2 \hat{j} + 3 \hat{k}$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

1 6

2 7

3

\sqrt{17}

4

\sqrt{14}

Explanation:

B Given,
Equation of line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ or
$\vec{r} = 6 i + 7 j + 7 k + λ (3 i + 2 j - 2 k)$
Where, $\vec{a} = 6 i + 7 j + 7 k$ and $\vec{b} = 3 i + 2 j - 2 k$
Let $\vec{c} = i + 2 j + 3 k$
$\Rightarrow Required distance = \frac{| (\vec{c} - \vec{a}) \times \vec{b} |}{| \vec{b} |}$
$= | \frac{(- 5 i - 5 j - 4 k) \times (3 i + 2 j - 2 k)}{\sqrt{17}} |$
$= \frac{| 18 i - 22 j + 5 k |}{\sqrt{17}} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{49}$
$\Rightarrow$ Length $(d) = 7$

AMU-2018

Three Dimensional Geometry

121375 Let $P$ be the point of intersection of the line $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2}$ and the plane $x + y + z = 2$ . If the distance of the point $P$ from the plane $3 x$ $- 4 y + 12 z = 32$ is $q$ , then $q$ and $2 q$ are the roots of the equation

1

x^{2} = 18 x - 72 = 0

2

x^{2} + 18 x + 72 = 0

3

x^{2} - 18 x + 72 = 0

4

x^{2} + 18 x - 72 = 0

Explanation:

C $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = k$
$∴ x = 3 k - 3, y = k - 2, z = 1 - 2 k$
Since, given that,
$P = (3 k - 3, k - 2, 1 - 2 k)$
Given, be the point of intersection of the line and the plane $x + y = z = 2$
$∴ (3 k - 3) + (k - 2) + (1 - 2 k) = 2$
$2 k - 4 = 2$
$k = 3$
Thus, $P = (6, 1, - 5)$
Now, distance point $P$ from the plane
$3 x - 4 y + 12 z = 32$
$= | \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} | = | \frac{- 78}{\sqrt{169}} |$
$q = | \frac{- 78}{13} | = \frac{78}{13} = 6$
Since, $q = 6$
So, $2 q = 12$
We know that,
roots, $α = q = 6$ and $β = 2 q = 12$
Now, quadratic equation-
$x^{2} - (α + β) x + α β = 0$
$\Rightarrow x^{2} - (12 + 6) x + 12 \times 6 = 0$
$\Rightarrow x^{2} - 18 x + 72 = 0$

JEE Main-10.04.2023

Three Dimensional Geometry

121378 Let the line $\frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5}$ intersect the lines $\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1}$ and $\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $A B$ from the plane $2 x - 2 y + z = 14$ is

1 4

2

\frac{10}{3}

3 3

4

\frac{11}{3}

Explanation:

A Given,
$l_{1} := \frac{x}{1} = \frac{y - 6}{- 2} = \frac{z + 8}{5} = λ$
$l_{2} := \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = μ$
$l_{3} := \frac{x + 3}{6} = \frac{y - 3}{- 3} = \frac{z - 6}{1} = γ$
Intersection of $l_{1}$ and $l_{2}$
$\Rightarrow (λ, - 2 λ + 6, 5 λ - 8) & (4 μ + 5, 3 μ + 7, μ - 2)$
$\Rightarrow λ = 1$ and $μ = - 1$
So, point A $(1, 4, - 3)$
Intersection of (i) and (iii)
$(λ, - 2 λ + 6, 5 λ - 8) & (6 γ - 3, - 3 γ + 3, γ + 6)$
$\Rightarrow λ = 3$ and $γ = 1$
So, point $(B) = (2, 2, 2)$
Perpendicular distance from plane $2 x - 2 y + z = 14$ is
$\Rightarrow d = | \frac{2 (2) - 2 (2) + 2 - 14}{\sqrt{4 + 4 + 1}} |$

JEE Main-10.04.2023

Three Dimensional Geometry

121380 A plane $E$ is perpendicular to the two planes $2 x$ $- 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $3 \sqrt{2}$ , then $(PQ)^{2}$ is equal to

1 9

2 12

3 21

4 33

Explanation:

C Given, equation of planes :
$P_{1} : 2 x - 2 y + z = 0$
$P_{2} : x - y + 2 z = 4$
\& Points: P $(1, - 1, 1), Q (a, a, 2)$
Equation of plane $E$ is
$a (x - 1) + b (y + 1) + c (z - 1) = 0$
Also, it is perpendicular to $P_{1}$ and $P_{2}$
$\Rightarrow 2 x - 2 b + c = 0$
And
$a - b + 2 c = 0$
On solving (i) and (ii)
$\frac{a}{- 4 + 1} = \frac{b}{1 - 4} = \frac{c}{0} = λ$
$\Rightarrow a = 3 λ, b = - 3 λ$ and $c = 0$
\& Equation of plane $E$ -
$- 3 λ (x - 1) - 3 λ (y + 1) + 0 (z - 1) = 0$
Now, distance from plane to $Q$ is -
$d = | \frac{1 \times a + 1 \times a + 0 \times 2}{\sqrt{1^{2} + 1^{2} + 0^{2}}} | = 3 \sqrt{2}$
$\Rightarrow \frac{2 a}{\sqrt{2}} = 3 \sqrt{2}$
$\Rightarrow a = 3$
$∴ Q = (3, 3, 2)$
$Now, (PQ)^{2} = ((3 - 1)^{2} + (3 + 1)^{2} + (2 - 1)^{2})$
$\Rightarrow (PQ)^{2} = 21$ ( $∴$ Given)

JEE Main-25.07.2022

Three Dimensional Geometry

121387 The length of perpendicular from point $\hat{i} + 2 \hat{j} + 3 \hat{k}$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

1 6

2 7

3

\sqrt{17}

4

\sqrt{14}

Explanation:

B Given,
Equation of line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ or
$\vec{r} = 6 i + 7 j + 7 k + λ (3 i + 2 j - 2 k)$
Where, $\vec{a} = 6 i + 7 j + 7 k$ and $\vec{b} = 3 i + 2 j - 2 k$
Let $\vec{c} = i + 2 j + 3 k$
$\Rightarrow Required distance = \frac{| (\vec{c} - \vec{a}) \times \vec{b} |}{| \vec{b} |}$
$= | \frac{(- 5 i - 5 j - 4 k) \times (3 i + 2 j - 2 k)}{\sqrt{17}} |$
$= \frac{| 18 i - 22 j + 5 k |}{\sqrt{17}} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{49}$
$\Rightarrow$ Length $(d) = 7$

AMU-2018

Three Dimensional Geometry

121375 Let $P$ be the point of intersection of the line $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2}$ and the plane $x + y + z = 2$ . If the distance of the point $P$ from the plane $3 x$ $- 4 y + 12 z = 32$ is $q$ , then $q$ and $2 q$ are the roots of the equation

1

x^{2} = 18 x - 72 = 0

2

x^{2} + 18 x + 72 = 0

3

x^{2} - 18 x + 72 = 0

4

x^{2} + 18 x - 72 = 0

Explanation:

C $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = k$
$∴ x = 3 k - 3, y = k - 2, z = 1 - 2 k$
Since, given that,
$P = (3 k - 3, k - 2, 1 - 2 k)$
Given, be the point of intersection of the line and the plane $x + y = z = 2$
$∴ (3 k - 3) + (k - 2) + (1 - 2 k) = 2$
$2 k - 4 = 2$
$k = 3$
Thus, $P = (6, 1, - 5)$
Now, distance point $P$ from the plane
$3 x - 4 y + 12 z = 32$
$= | \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} | = | \frac{- 78}{\sqrt{169}} |$
$q = | \frac{- 78}{13} | = \frac{78}{13} = 6$
Since, $q = 6$
So, $2 q = 12$
We know that,
roots, $α = q = 6$ and $β = 2 q = 12$
Now, quadratic equation-
$x^{2} - (α + β) x + α β = 0$
$\Rightarrow x^{2} - (12 + 6) x + 12 \times 6 = 0$
$\Rightarrow x^{2} - 18 x + 72 = 0$

JEE Main-10.04.2023

Three Dimensional Geometry

121378 Let the line $\frac{x}{1} = \frac{6 - y}{2} = \frac{z + 8}{5}$ intersect the lines $\frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1}$ and $\frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $A B$ from the plane $2 x - 2 y + z = 14$ is

1 4

2

\frac{10}{3}

3 3

4

\frac{11}{3}

Explanation:

A Given,
$l_{1} := \frac{x}{1} = \frac{y - 6}{- 2} = \frac{z + 8}{5} = λ$
$l_{2} := \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = μ$
$l_{3} := \frac{x + 3}{6} = \frac{y - 3}{- 3} = \frac{z - 6}{1} = γ$
Intersection of $l_{1}$ and $l_{2}$
$\Rightarrow (λ, - 2 λ + 6, 5 λ - 8) & (4 μ + 5, 3 μ + 7, μ - 2)$
$\Rightarrow λ = 1$ and $μ = - 1$
So, point A $(1, 4, - 3)$
Intersection of (i) and (iii)
$(λ, - 2 λ + 6, 5 λ - 8) & (6 γ - 3, - 3 γ + 3, γ + 6)$
$\Rightarrow λ = 3$ and $γ = 1$
So, point $(B) = (2, 2, 2)$
Perpendicular distance from plane $2 x - 2 y + z = 14$ is
$\Rightarrow d = | \frac{2 (2) - 2 (2) + 2 - 14}{\sqrt{4 + 4 + 1}} |$

JEE Main-10.04.2023

Three Dimensional Geometry

121380 A plane $E$ is perpendicular to the two planes $2 x$ $- 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $3 \sqrt{2}$ , then $(PQ)^{2}$ is equal to

1 9

2 12

3 21

4 33

Explanation:

C Given, equation of planes :
$P_{1} : 2 x - 2 y + z = 0$
$P_{2} : x - y + 2 z = 4$
\& Points: P $(1, - 1, 1), Q (a, a, 2)$
Equation of plane $E$ is
$a (x - 1) + b (y + 1) + c (z - 1) = 0$
Also, it is perpendicular to $P_{1}$ and $P_{2}$
$\Rightarrow 2 x - 2 b + c = 0$
And
$a - b + 2 c = 0$
On solving (i) and (ii)
$\frac{a}{- 4 + 1} = \frac{b}{1 - 4} = \frac{c}{0} = λ$
$\Rightarrow a = 3 λ, b = - 3 λ$ and $c = 0$
\& Equation of plane $E$ -
$- 3 λ (x - 1) - 3 λ (y + 1) + 0 (z - 1) = 0$
Now, distance from plane to $Q$ is -
$d = | \frac{1 \times a + 1 \times a + 0 \times 2}{\sqrt{1^{2} + 1^{2} + 0^{2}}} | = 3 \sqrt{2}$
$\Rightarrow \frac{2 a}{\sqrt{2}} = 3 \sqrt{2}$
$\Rightarrow a = 3$
$∴ Q = (3, 3, 2)$
$Now, (PQ)^{2} = ((3 - 1)^{2} + (3 + 1)^{2} + (2 - 1)^{2})$
$\Rightarrow (PQ)^{2} = 21$ ( $∴$ Given)

JEE Main-25.07.2022

Three Dimensional Geometry

121387 The length of perpendicular from point $\hat{i} + 2 \hat{j} + 3 \hat{k}$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is

1 6

2 7

3

\sqrt{17}

4

\sqrt{14}

Explanation:

B Given,
Equation of line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ or
$\vec{r} = 6 i + 7 j + 7 k + λ (3 i + 2 j - 2 k)$
Where, $\vec{a} = 6 i + 7 j + 7 k$ and $\vec{b} = 3 i + 2 j - 2 k$
Let $\vec{c} = i + 2 j + 3 k$
$\Rightarrow Required distance = \frac{| (\vec{c} - \vec{a}) \times \vec{b} |}{| \vec{b} |}$
$= | \frac{(- 5 i - 5 j - 4 k) \times (3 i + 2 j - 2 k)}{\sqrt{17}} |$
$= \frac{| 18 i - 22 j + 5 k |}{\sqrt{17}} = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{49}$
$\Rightarrow$ Length $(d) = 7$

AMU-2018