Let $P$ be a plane passing through the points $(2,1,0)$, $(4,1,1)$ and $(5,0,1)$ and $R$ be any point $(2,1,6)$. Then the image of $R$ in the plane $P$ is
(1) $(6,5,2)$
(2) $(6,5,-2)$
(3) $(4,3,2)$
(4) $(3,4,-2)$

The plane passing through the points $(1, 2, 1)$, $(2, 1, 2)$ and parallel to the line, $2x = 3y, z = 1$ also passes through the point
(1) $(0, 6, -2)$
(2) $(-2, 0, 1)$
(3) $(0, -6, 2)$
(4) $(2, 0, -1)$

The distance of the point $( 1 , - 2,3 )$ from the plane $x - y + z = 5$ measured parallel to the line $\frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { - 6 }$ is:
(1) $\frac { 7 } { 5 }$
(2) 1
(3) $\frac { 1 } { 7 }$
(4) 7

If $( a , b , c )$ is the image of the point $( 1,2 , - 3 )$ in the line, $\frac { x + 1 } { 2 } = \frac { y - 3 } { - 2 } = \frac { z } { - 1 }$, then $a + b + c$ is equal to:
(1) 2
(2) - 1
(3) 3
(4) 1

If for some $\alpha \in \mathrm{R}$, the lines $L_1: \frac{x+1}{2} = \frac{y-2}{-1} = \frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha} = \frac{y+1}{5-\alpha} = \frac{z+1}{1}$ are coplanar, then the line $L_2$ passes through the point:
(1) $(10, 2, 2)$
(2) $(2, -10, -2)$
(3) $(10, -2, -2)$
(4) $(-2, 10, 2)$

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and $l ^ { 2 } + m ^ { 2 } - n ^ { 2 } = 0$. Then the value of $\sin ^ { 4 } \alpha + \cos ^ { 4 } \alpha$ is :
(1) $\frac { 5 } { 8 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 3 } { 8 }$
(4) $\frac { 3 } { 4 }$

The lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2 , ( ab \neq 0 )$ are coplanar, if:
(1) $b = 1 , a \in R - \{ 0 \}$
(2) $a = 1 , b \in R - \{ 0 \}$
(3) $a = 2 , b = 2$
(4) $a = 2 , b = 3$

The equation of the plane passing through the line of intersection of the planes $\vec { r } \cdot ( \hat { i } + \hat { j } + \widehat { k } ) = 1$ and $\vec { r } \cdot ( 2 \hat { i } + 3 \hat { j } - \hat { k } ) + 4 = 0$ and parallel to the $x$-axis, is (1) $\vec { r } \cdot ( \hat { i } + 3 \widehat { k } ) + 6 = 0$ (2) $\vec { r } \cdot ( \hat { i } - 3 \widehat { k } ) + 6 = 0$ (3) $\vec { r } \cdot ( \hat { j } - 3 \widehat { k } ) - 6 = 0$ (4) $\vec { r } \cdot ( \hat { j } - 3 \widehat { k } ) + 6 = 0$

If the equation of plane passing through the mirror image of a point $( 2,3,1 )$ with respect to line $\frac { x + 1 } { 2 } = \frac { y - 3 } { 1 } = \frac { z + 2 } { - 1 }$ and containing the line $\frac { x - 2 } { 3 } = \frac { 1 - y } { 2 } = \frac { z + 1 } { 1 }$ is $\alpha x + \beta y + \gamma z = 24$ then $\alpha + \beta + \gamma$ is equal to:
(1) 20
(2) 19
(3) 18
(4) 21

Consider the line $L$ given by the equation $\frac { x - 3 } { 2 } = \frac { y - 1 } { 1 } = \frac { z - 2 } { 1 }$. Let $Q$ be the mirror image of the point $( 2,3 , - 1 )$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?
(1) $( - 1,1,2 )$
(2) $( 1,1,1 )$
(3) $( 1,1,2 )$
(4) $( 1,2,2 )$

Let the equation of the plane, that passes through the point $( 1,4 , - 3 )$ and contains the line of intersection of the planes $3 x - 2 y + 4 z - 7 = 0$ and $x + 5 y - 2 z + 9 = 0$, be $\alpha x + \beta y + \gamma z + 3 = 0$, then $\alpha + \beta + \gamma$ is equal to :
(1) $- 15$
(2) 15
(3) $- 23$
(4) 23

Let $P$ be a plane $l x + m y + n z = 0$ containing the line, $\frac { 1 - x } { 1 } = \frac { y + 4 } { 2 } = \frac { z + 2 } { 3 }$. If plane $P$ divides the line segment $A B$ joining points $A ( - 3 , - 6,1 )$ and $B ( 2,4 , - 3 )$ in ratio $k : 1$ then the value of $k$ is

Let the foot of the perpendicular from the point $( 1,2,4 )$ on the line $\frac { x + 2 } { 4 } = \frac { y - 1 } { 2 } = \frac { z + 1 } { 3 }$ be $P$. Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$ is
(1) $\frac { 50 } { 13 }$
(2) $\frac { 63 } { 13 }$
(3) $\frac { 65 } { 13 }$
(4) 4

Let $P$ be the plane containing the straight line $\frac { x - 3 } { 9 } = \frac { y + 4 } { - 1 } = \frac { z - 7 } { - 5 }$ and perpendicular to the plane containing the straight lines $\frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { 5 }$ and $\frac { x } { 3 } = \frac { y } { 7 } = \frac { z } { 8 }$. If $d$ is the distance of $P$ from the point $(2,-5,11)$, then $d ^ { 2 }$ is equal to
(1) $\frac { 147 } { 2 }$
(2) 96
(3) $\frac { 32 } { 3 }$
(4) 54

If the line of intersection of the planes $a x + b y = 3$ and $a x + b y + c z = 0 , a > 0$ makes an angle $30 ^ { \circ }$ with the plane $y - z + 2 = 0$, then the direction cosines of the line are
(1) $\frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } , 0$
(2) $\frac { 1 } { \sqrt { 2 } } , \frac { - 1 } { \sqrt { 2 } } , 0$
(3) $\frac { 1 } { \sqrt { 5 } } , - \frac { 2 } { \sqrt { 5 } } , 0$
(4) $\frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } , 0$

Let the lines $\frac { x - 1 } { \lambda } = \frac { y - 2 } { 1 } = \frac { z - 3 } { 2 }$ and $\frac { x + 26 } { - 2 } = \frac { y + 18 } { 3 } = \frac { z + 28 } { \lambda }$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lie on $P$?
(1) $( 0 , - 2 , - 2 )$
(2) $( - 5,0 , - 1 )$
(3) $( 3 , - 1,0 )$
(4) $( 0,4,5 )$

The shortest distance between the lines $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 1 } { - 1 }$ and $\frac { x + 3 } { 2 } = \frac { y - 6 } { 1 } = \frac { z - 5 } { 3 }$ is
(1) $\frac { 18 } { \sqrt { 5 } }$
(2) $\frac { 22 } { 3 \sqrt { 5 } }$
(3) $\frac { 46 } { 3 \sqrt { 5 } }$
(4) $6 \sqrt { 3 }$

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x + k y - 5 z = 1$ and $3 k x - k y + z = 5 , k < 3$ and intercepts a unit length on positive $x$-axis, then the intercept made by the plane $P$ on the $y$-axis is
(1) $\frac { 1 } { 11 }$
(2) $\frac { 5 } { 11 }$
(3) 6
(4) 7

A plane $P$ is parallel to two lines whose direction ratios are $- 2,1 , - 3$ and $- 1,2 , - 2$ and it contains the point $( 2,2 , - 2 )$. Let $P$ intersect the co-ordinate axes at the points $A , B , C$ making the intercepts $\alpha , \beta , \gamma$. If $V$ is the volume of the tetrahedron $O A B C$, where $O$ is the origin and $p = \alpha + \beta + \gamma$, then the ordered pair $( V , p )$ is equal to
(1) $( 48 , - 13 )$
(2) $( 24 , - 13 )$
(3) $( 48,11 )$
(4) $( 24 , - 5 )$

The line of shortest distance between the lines $\frac { x - 2 } { 0 } = \frac { y - 1 } { 1 } = \frac { z } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 5 } { 2 } = \frac { z - 1 } { 1 }$ makes an angle of $\sin ^ { - 1 } \sqrt { \frac { 2 } { 27 } }$ with the plane $P : ax - y - z = 0$, $a > 0$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta - \gamma$ is equal to $\_\_\_\_$.

Let the line $\frac { x - 3 } { 7 } = \frac { y - 2 } { - 1 } = \frac { z - 3 } { - 4 }$ intersect the plane containing the lines $\frac { x - 4 } { 1 } = \frac { y + 1 } { - 2 } = \frac { z } { 1 }$ and $4 a x - y + 5 z - 7 a = 0 = 2 x - 5 y - z - 3 , a \in \mathbb { R }$ at the point $P ( \alpha , \beta , \gamma )$. Then the value of $\alpha + \beta + \gamma$ equals $\_\_\_\_$ .

Let $P$ be the plane passing through the intersection of the planes $\vec{r}\cdot(\hat{i}+3\hat{j}-\hat{k}) = 5$ and $\vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k}) = 3$, and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points $X$ and $Y$ with respect to the plane $P$ are
(1) on the same side
(2) on opposite sides
(3) $X$ lies on $P$
(4) $Y$ lies on $P$

The plane, passing through the points $( 0 , - 1 , 2 )$ and $( - 1 , 2 , 1 )$ and parallel to the line passing through $( 5 , 1 , - 7 )$ and $( 1 , - 1 , - 1 )$, also passes through the point
(1) $( - 2 , 5 , 0 )$
(2) $( 1 , - 2 , 1 )$
(3) $( 2 , 0 , 1 )$
(4) $( 0 , 5 , - 2 )$

The line, that is coplanar to the line $\frac { x + 3 } { - 3 } = \frac { y - 1 } { 1 } = \frac { z - 5 } { 5 }$, is
(1) $\frac { x + 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 4 }$
(2) $\frac { x + 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$
(3) $\frac { x - 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$
(4) $\frac { x + 1 } { 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$

Let the foot of perpendicular of the point $P ( 3 , - 2 , - 9 )$ on the plane passing through the points $( - 1 , - 2 , - 3 ) , ( 9,3,4 ) , ( 9 , - 2,1 )$ be $Q ( \alpha , \beta , \gamma )$. Then the distance of $Q$ from the origin is
(1) $\sqrt { 42 }$
(2) $\sqrt { 38 }$
(3) $\sqrt { 35 }$
(4) $\sqrt { 29 }$