Let $\ell _ { 1 }$ and $\ell _ { 2 }$ be the lines $\vec { r } _ { 1 } = \lambda ( \hat { i } + \hat { j } + \hat { k } )$ and $\vec { r } _ { 2 } = ( \hat { j } - \hat { k } ) + \mu ( \hat { i } + \hat { k } )$, respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell _ { 1 }$. For a plane $H$, let $d ( H )$ denote the smallest possible distance between the points of $\ell _ { 2 }$ and $H$. Let $H _ { 0 }$ be a plane in $X$ for which $d \left( H _ { 0 } \right)$ is the maximum value of $d ( H )$ as $H$ varies over all planes in $X$.
Match each entry in List-I to the correct entries in List-II.
List-I
(P) The value of $d \left( H _ { 0 } \right)$ is
(Q) The distance of the point $( 0,1,2 )$ from $H _ { 0 }$ is
(R) The distance of origin from $H _ { 0 }$ is
(S) The distance of origin from the point of intersection of planes $y = z , x = 1$ and $H _ { 0 }$ is
List-II
(1) $\sqrt { 3 }$
(2) $\frac { 1 } { \sqrt { 3 } }$
(3) 0
(4) $\sqrt { 2 }$
(5) $\frac { 1 } { \sqrt { 2 } }$
The correct option is:
(A) $( P ) \rightarrow ( 2 )$ $( Q ) \rightarrow ( 4 )$ $( R ) \rightarrow ( 5 )$ $( S ) \rightarrow ( 1 )$
(B) $( P ) \rightarrow ( 5 )$ $( Q ) \rightarrow ( 4 )$ $( R ) \rightarrow ( 3 )$ $( S ) \rightarrow ( 1 )$
(C) $( P ) \rightarrow ( 2 )$ $( Q ) \rightarrow ( 1 )$ $( R ) \rightarrow ( 3 )$ $( S ) \rightarrow ( 2 )$
(D) $( P ) \rightarrow ( 5 )$ $( Q ) \rightarrow ( 1 )$ $( R ) \rightarrow ( 4 )$ $( S ) \rightarrow ( 2 )$

Let $\gamma \in \mathbb { R }$ be such that the lines $L _ { 1 } : \frac { x + 11 } { 1 } = \frac { y + 21 } { 2 } = \frac { z + 29 } { 3 }$ and $L _ { 2 } : \frac { x + 16 } { 3 } = \frac { y + 11 } { 2 } = \frac { z + 4 } { \gamma }$ intersect. Let $R _ { 1 }$ be the point of intersection of $L _ { 1 }$ and $L _ { 2 }$. Let $O = ( 0,0,0 )$, and $\hat { n }$ denote a unit normal vector to the plane containing both the lines $L _ { 1 }$ and $L _ { 2 }$.
Match each entry in List-I to the correct entry in List-II.
List-I
(P) $\gamma$ equals
(Q) A possible choice for $\hat { n }$ is
(R) $\overrightarrow { O R _ { 1 } }$ equals
(S) A possible value of $\overrightarrow { O R _ { 1 } } \cdot \hat { n }$ is
List-II
(1) $- \hat { i } - \hat { j } + \hat { k }$
(2) $\sqrt { \frac { 3 } { 2 } }$
(3) 1
(4) $\frac { 1 } { \sqrt { 6 } } \hat { i } - \frac { 2 } { \sqrt { 6 } } \hat { j } + \frac { 1 } { \sqrt { 6 } } \hat { k }$
(5) $\sqrt { \frac { 2 } { 3 } }$
The correct option is
(A) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (2)$
(B) $(\mathrm{P}) \rightarrow (5)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (2)$
(C) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (5)$
(D) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (1)$, $(\mathrm{R}) \rightarrow (4)$, $(\mathrm{S}) \rightarrow (5)$

Let $L$ be the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$. If $L$ makes an angles $\alpha$ with the positive $x$-axis, then $\cos \alpha$ equals
(1) $\frac { 1 } { \sqrt { 3 } }$
(2) $\frac { 1 } { 2 }$
(3) 1
(4) $\frac { 1 } { \sqrt { 2 } }$

If a line makes an angle of $\frac { \pi } { 4 }$ with the positive directions of each of $x$-axis and $y$-axis, then the angle that the line makes with the positive direction of the $z$-axis is
(1) $\frac { \pi } { 6 }$
(2) $\frac { \pi } { 3 }$
(3) $\frac { \pi } { 4 }$
(4) $\frac { \pi } { 2 }$

If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to
(1) $\frac{2}{9}$
(2) $\frac{9}{2}$
(3) 0
(4) $-1$

If the lines $\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k}$ and $\frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1}$ are coplanar, then $k$ can be
(1) $-1$ or $-3$
(2) $-1$ or $3$
(3) $1$ or $-1$
(4) $0$ or $-3$

The image of the line $\frac { x - 1 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 4 } { - 5 }$ in the plane $2 x - y + z + 3 = 0$ is the line
(1) $\frac { x - 3 } { 3 } = \frac { y + 5 } { 1 } = \frac { z - 2 } { - 5 }$
(2) $\frac { x - 3 } { - 3 } = \frac { y + 5 } { - 1 } = \frac { z - 2 } { 5 }$
(3) $\frac { x + 3 } { 3 } = \frac { y - 5 } { 1 } = \frac { z - 2 } { - 5 }$
(4) $\frac { x + 3 } { - 3 } = \frac { y - 5 } { - 1 } = \frac { z + 2 } { 5 }$

Equation of the plane which passes through the point of intersection of lines $\frac { x - 1 } { 3 } = \frac { y - 2 } { 1 } = \frac { z - 3 } { 2 }$ and $\frac { x - 3 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$ and has the largest distance from the origin is:
(1) $4 x + 3 y + 5 z = 50$
(2) $3 x + 4 y + 5 z = 49$
(3) $5 x + 4 y + 3 z = 57$
(4) $7 x + 2 y + 4 z = 54$

If the angle between the line $2 ( x + 1 ) = y = z + 4$ and the plane $2 x - y + \sqrt { \lambda } z + 4 = 0$ is $\frac { \pi } { 6 }$, then the value of $\lambda$ is
(1) $\frac { 45 } { 7 }$
(2) $\frac { 135 } { 11 }$
(3) $\frac { 135 } { 7 }$
(4) $\frac { 45 } { 11 }$

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l ^ { 2 } = m ^ { 2 } + n ^ { 2 }$ is
(1) $\frac { \pi } { 6 }$
(2) $\frac { \pi } { 2 }$
(3) $\frac { \pi } { 3 }$
(4) $\frac { \pi } { 4 }$

A line in the 3-dimensional space makes an angle $\theta \left( 0 < \theta \leq \frac { \pi } { 2 } \right)$ with both the $X$ and $Y$-axes. Then, the set of all values of $\theta$ is in the interval:
(1) $\left( \frac { \pi } { 3 } , \frac { \pi } { 2 } \right]$
(2) $\left( 0 , \frac { \pi } { 4 } \right]$
(3) $\left[ \frac { \pi } { 4 } , \frac { \pi } { 2 } \right]$
(4) $\left[ \frac { \pi } { 6 } , \frac { \pi } { 3 } \right]$

Equation of the line of the shortest distance between the lines $\frac { x } { 1 } = \frac { y } { - 1 } = \frac { z } { 1 }$ and $\frac { x - 1 } { 0 } = \frac { y + 1 } { - 2 } = \frac { z } { 1 }$ is
(1) $\frac { x } { - 2 } = \frac { y } { 1 } = \frac { z } { 2 }$
(2) $\frac { x } { 1 } = \frac { y } { - 1 } = \frac { z } { - 2 }$
(3) $\frac { x - 1 } { 1 } = \frac { y + 1 } { - 1 } = \frac { z } { - 2 }$
(4) $\frac { x - 1 } { 1 } = \frac { y + 1 } { - 1 } = \frac { z } { 1 }$

The equation of the plane containing the line $2x - 5y + z = 3$; $x + y + 4z = 5$, and parallel to the plane $x + 3y + 6z = 1$, is:
(1) $2x + 6y + 12z = 13$
(2) $x + 3y + 6z = -7$
(3) $x + 3y + 6z = 7$
(4) $2x + 6y + 12z = -13$

The distance of the point $( 1,0,2 )$ from the point of intersection of the line $\frac { x - 2 } { 3 } = \frac { y + 1 } { 4 } = \frac { z - 2 } { 12 }$ and the plane $x - y + z = 16$, is
(1) 13
(2) $2 \sqrt { 14 }$
(3) 8
(4) $3 \sqrt { 21 }$

The equation of the plane containing the line of intersection of $2 x - 5 y + z = 3 ; x + y + 4 z = 5$, and parallel to the plane, $x + 3 y + 6 z = 1$, is
(1) $2 x + 6 y + 12 z = - 13$
(2) $2 x + 6 y + 12 z = 13$
(3) $x + 3 y + 6 z = - 7$
(4) $x + 3 y + 6 z = 7$

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is:
(1) $3\sqrt{10}$
(2) $10\sqrt{3}$
(3) $\frac{10}{\sqrt{3}}$
(4) $\frac{20}{3}$

If the line $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$ lies in the plane $lx + my - z = 9$, then $l^2 + m^2$ is equal to:
(1) $26$
(2) $18$
(3) $1$
(4) $2$

If the line $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$ lies in the plane $lx + my - z = 9$, then $l^2 + m^2$ is equal to: (1) 18 (2) 5 (3) 2 (4) 26

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is: (1) $3\sqrt{10}$ (2) $10\sqrt{3}$ (3) $\frac{10}{\sqrt{3}}$ (4) $\frac{20}{3}$

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $$\frac { x - 1 } { 1 } = \frac { y + 2 } { - 2 } = \frac { z - 4 } { 3 } \quad \text{and} \quad \frac { x - 2 } { 2 } = \frac { y + 1 } { - 1 } = \frac { z + 7 } { - 1 }$$ is:
(1) $\frac { 10 } { \sqrt { 74 } }$
(2) $\frac { 20 } { \sqrt { 74 } }$
(3) $\frac { 5 } { \sqrt { 83 } }$
(4) $\frac { 10 } { \sqrt { 83 } }$

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\dfrac{x}{1} = \dfrac{y}{4} = \dfrac{z}{5}$ is $Q$, then $PQ$ is equal to:
(1) $3\sqrt{5}$
(2) $2\sqrt{42}$
(3) $\sqrt{42}$
(4) $6\sqrt{5}$

The coordinates of the foot of the perpendicular from the point $( 1 , - 2,1 )$ on the plane containing the lines $\frac { x + 1 } { 6 } = \frac { y - 1 } { 7 } = \frac { z - 3 } { 8 }$ and $\frac { x - 1 } { 3 } = \frac { y - 2 } { 5 } = \frac { z - 3 } { 7 }$, is:
(1) $( 2 , - 4,2 )$
(2) $( 1,1,1 )$
(3) $( 0,0,0 )$
(4) $( - 1,2 , - 1 )$

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}$, is:
(1) $\dfrac{20}{\sqrt{74}}$
(2) $\dfrac{10}{\sqrt{83}}$
(3) $\dfrac{5}{\sqrt{83}}$
(4) $\dfrac{10}{\sqrt{74}}$

The line of intersection of the planes $\vec { r } \cdot ( 3 \hat { i } - \hat { j } + \widehat { k } ) = 1$ and $\vec { r } \cdot ( \hat { i } + 4 \hat { j } - 2 \widehat { k } ) = 2$, is,
(1) $\frac { x - \frac { 6 } { 13 } } { 2 } = \frac { y - \frac { 5 } { 13 } } { 7 } = \frac { z } { - 13 }$
(2) $\frac { x - \frac { 4 } { 7 } } { 2 } = \frac { y } { - 7 } = \frac { z + \frac { 5 } { 7 } } { 13 }$
(3) $\frac { x - \frac { 6 } { 13 } } { 2 } = \frac { y - \frac { 5 } { 13 } } { - 7 } = \frac { z } { - 13 }$
(4) $\frac { x - \frac { 4 } { 7 } } { - 2 } = \frac { y } { 7 } = \frac { z - \frac { 5 } { 7 } } { 13 }$

If $L _ { 1 }$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0 , x - y + z + 1 = 0$ and $L _ { 2 }$ is the line of intersection of the planes $x + 2 y - z - 3 = 0,3 x - y + 2 z - 1 = 0$, then the distance of the origin from the plane, containing the lines $L _ { 1 }$ and $L _ { 2 }$ is
(1) $\frac { 1 } { \sqrt { 2 } }$
(2) $\frac { 1 } { 4 \sqrt { 2 } }$
(3) $\frac { 1 } { 3 \sqrt { 2 } }$
(4) $\frac { 1 } { 2 \sqrt { 2 } }$