A variable plane passes through a fixed point $( 3,2,1 )$ and meets $x , y$ and $z$-axes at $A , B \& C$ respectively. A plane is drawn parallel to the $yz$-plane through $A$, a second plane is drawn parallel to the $zx$-plane through $B$ and a third plane is drawn parallel to the $xy$-plane through $C$. Then the locus of the point of intersection of these three planes, is
(1) $\frac { 3 } { x } + \frac { 2 } { y } + \frac { 1 } { z } = 1$
(2) $\frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } = \frac { 11 } { 6 }$
(3) $x + y + z = 6$
(4) $\frac { x } { 3 } + \frac { y } { 2 } + \frac { z } { 1 } = 1$

A variable plane passes through a fixed point ( $3,2,1$ ) and meets $x , y$ and $z$ axes at $A , B$ and $C$ respectively. A plane is drawn parallel to $y z$ - plane through $A$, a second plane is drawn parallel $z x$ plane through $B$ and a third plane is drawn parallel to $x y$ - plane through $C$. Then the locus of the point of intersection of these three planes, is
(1) $( x + y + z = 6 )$
(2) $\frac { x } { 3 } + \frac { y } { 2 } + \frac { z } { 1 } = 1$
(3) $\frac { 3 } { x } + \frac { 2 } { y } + \frac { 1 } { z } = 1$
(4) $\frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } = \frac { 11 } { 6 }$

An angle between the plane $x + y + z = 5$ and the line of intersection of the planes, $3 x + 4 y + z - 1 = 0$ and $5 x + 8 y + 2 z + 14 = 0$ is
(1) $\cos ^ { - 1 } \left( \sqrt { \frac { 3 } { 17 } } \right)$
(2) $\cos ^ { - 1 } \left( \frac { 3 } { \sqrt { 17 } } \right)$
(3) $\sin ^ { - 1 } \left( \frac { 3 } { \sqrt { 17 } } \right)$
(4) $\sin ^ { - 1 } \left( \sqrt { \frac { 3 } { 17 } } \right)$

An angle between the plane, $x + y + z = 5$ and the line of intersection of the planes, $3 x + 4 y + z - 1 = 0$ and $5 x + 8 y + 2 z + 14 = 0$, is
(1) $\cos ^ { - 1 } \left( \frac { 3 } { \sqrt { 17 } } \right)$
(2) $\cos ^ { - 1 } \left( \sqrt { \frac { 3 } { 17 } } \right)$
(3) $\sin ^ { - 1 } \left( \frac { 3 } { \sqrt { 17 } } \right)$
(4) $\sin ^ { - 1 } \left( \sqrt { \frac { 3 } { 17 } } \right)$

If an angle between the line, $\frac { x + 1 } { 2 } = \frac { y - 2 } { 1 } = \frac { z - 3 } { - 2 }$ and the plane, $x - 2 y - k z = 3$ is $\cos ^ { - 1 } \left( \frac { 2 \sqrt { 2 } } { 3 } \right)$, then a value of $k$ is
(1) $\sqrt { \frac { 5 } { 3 } }$
(2) $\sqrt { \frac { 3 } { 5 } }$
(3) $- \frac { 3 } { 5 }$
(4) $- \frac { 5 } { 3 }$

The magnitude of the projection of the vector $2\hat{i} + 3\hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 3\hat{k}$, is:
(1) $3\sqrt{6}$
(2) $\sqrt{\frac{3}{2}}$
(3) $\sqrt{6}$
(4) $\frac{\sqrt{3}}{2}$

If the lines $x = ay + b,\, z = cy + d$ and $x = a'z + b',\, y = c'z + d'$ are perpendicular, then
(1) $cc' + a + a' = 0$
(2) $aa' + c + c' = 0$
(3) $bb' + cc' + 1 = 0$
(4) $ab' + bc' + 1 = 0$

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $\left( - \lambda ^ { 2 } , 1,1 \right) , \left( 1 , - \lambda ^ { 2 } , 1 \right)$ and $\left( 1,1 , - \lambda ^ { 2 } \right)$ also passes through the point $( - 1 , - 1,1 )$. Then $S$ is equal to :
(1) $\{ \sqrt { 3 } \}$
(2) $\{ 3 , - 3 \}$
(3) $\{ 1 , - 1 \}$
(4) $\{ \sqrt { 3 } , - \sqrt { 3 } \}$

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is
(1) $\vec { r } \times ( \hat { i } + \hat { k } ) + 2 = 0$
(2) $\vec { r } \cdot ( \hat { i } - \hat { k } ) - 2 = 0$
(3) $\vec { r } \times ( \hat { i } - \hat { k } ) + 2 = 0$
(4) $\vec { r } \cdot ( \hat { i } - \hat { k } ) + 2 = 0$

The length of the perpendicular from the point $(2,-1,4)$ on the straight line $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$ is
(1) greater than 3 but less than 4
(2) greater than 4
(3) less than 2
(4) greater than 2 but less than 3

A plane passing though the points $( 0 , - 1,0 )$ and $( 0,0,1 )$ and making an angle $\frac { \pi } { 4 }$ with the plane $y - z + 5 = 0$, also passes through the point
(1) $( \sqrt { 2 } , - 1,4 )$
(2) $( \sqrt { 2 } , 1,4 )$
(3) $( - \sqrt { 2 } , - 1 , - 4 )$
(4) $( - \sqrt { 2 } , 1 , - 4 )$

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:
(1) $3x + 2y - 3z = 0$
(2) $x + 2y - 2z = 0$
(3) $x - 2y + z = 0$
(4) $5x + 2y - 4z = 0$

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1,1,0)$ is
(1) $x - 3y - 2z = -2$
(2) $x + 3y + z = 4$
(3) $x - y - z = 0$
(4) $2x - z = 2$

If the line, $\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 2 } { 4 }$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$, then the distance of $P$ from the origin is,
(1) $2 \sqrt { 5 }$
(2) $\frac { 9 } { 2 }$
(3) $\frac { \sqrt { 5 } } { 2 }$
(4) $\frac { 7 } { 2 }$

If for some $\alpha$ and $\beta$ in $R$, the intersection of the following three planes $x + 4 y - 2 z = 1$ $x + 7 y - 5 z = \beta$ $x + 5 y + \alpha z = 5$ is a line in $R ^ { 3 }$, then $\alpha + \beta$ is equal to:
(1) 0
(2) 10
(3) 2
(4) - 10

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 lines $\vec { r } = ( \hat { i } - \hat { j } ) + l ( 2 \hat { i } + \widehat { k } )$ and $\vec { r } = ( 2 \hat { i } - \hat { j } ) + m ( \hat { i } + \hat { j } - \widehat { k } )$
(1) Do not intersect for any values of $l$ and $m$
(2) Intersect for all values of $l$ and $m$
(3) Intersect when $l = 2$ and $m = \frac { 1 } { 2 }$
(4) Intersect when $l = 1$ and $m = 2$

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) $(-2, 10, 2)$

The equation of the line through the point $( 0,1,2 )$ and perpendicular to the line $\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 1 } { - 2 }$ is :
(1) $\frac { x } { 3 } = \frac { y - 1 } { - 4 } = \frac { z - 2 } { 3 }$
(2) $\frac { x } { 3 } = \frac { y - 1 } { 4 } = \frac { z - 2 } { 3 }$
(3) $\frac { x } { - 3 } = \frac { y - 1 } { 4 } = \frac { z - 2 } { 3 }$
(4) $\frac { x } { 3 } = \frac { y - 1 } { 4 } = \frac { z - 2 } { - 3 }$

A plane passes through the points $A ( 1,2,3 ) , B ( 2,3,1 )$ and $C ( 2,4,2 )$. If $O$ is the origin and $P$ is $( 2 , - 1,1 )$, then the projection of $\overrightarrow { O P }$ on this plane is of length:
(1) $\sqrt { \frac { 2 } { 5 } }$
(2) $\sqrt { \frac { 2 } { 7 } }$
(3) $\sqrt { \frac { 2 } { 3 } }$
(4) $\sqrt { \frac { 2 } { 11 } }$

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) $( 1,2,2 )$