In coordinate space, the $x$-coordinate of the point where the plane passing through the point $( 2,0,5 )$ and containing the line $x - 1 = 2 - y = \frac { z + 1 } { 2 }$ meets the $x$-axis is? [3 points]
(1) $\frac { 9 } { 2 }$
(2) 4
(3) $\frac { 7 } { 2 }$
(4) 3
(5) $\frac { 5 } { 2 }$

The plane $E : 3 x _ { 1 } + 2 x _ { 2 } + 2 x _ { 3 } = 6$ contains a point whose three coordinates are equal. Determine these coordinates.

134- From point $A(5,-2,1)$, a line perpendicular to the plane with equation $x = t+1$, $y = -2t+1$, $z = 2t-3$ is drawn. What are the coordinates of the intersection point of this line and the plane?
(1) $(2,-1,-1)$ (2) $(1,1,-3)$ (3) $(4,5,3)$ (4) $(3,-3,1)$

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) $3 \sqrt { 21 }$

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) $\frac { 7 } { 2 }$

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 $\_\_\_\_$ .

The plane $2 x - y + z = 4$ intersects the line segment joining the points $\mathrm { A } ( \mathrm { a } , - 2,4 )$ and $\mathrm { B } ( 2 , \mathrm {~b} , - 3 )$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $\sqrt { 5 }$. If $a b < 0$ and P is the point $( \mathrm { a } - \mathrm { b } , \mathrm { b } , 2 \mathrm {~b} - \mathrm { a } )$ then $\mathrm { CP } ^ { 2 }$ is equal to: (1) $\frac { 17 } { 3 }$ (2) $\frac { 16 } { 3 }$ (3) $\frac { 73 } { 3 }$ (4) $\frac { 97 } { 3 }$

If the lines $\frac { x - 1 } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 1 }$ and $\frac { x - a } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 3 } { 1 }$ intersects at the point $P$, then the distance of the point $P$ from the plane $z = a$ is : (1) 16 (2) 28 (3) 10 (4) 22

Let $P Q R$ be a triangle with $R ( - 1,4,2 )$. Suppose $M ( 2,1,2 )$ is the mid point of $P Q$. The distance of the centroid of $\triangle P Q R$ from the point of intersection of the line $\frac { x - 2 } { 0 } = \frac { y } { 2 } = \frac { z + 3 } { - 1 }$ and $\frac { x - 1 } { 1 } = \frac { y + 3 } { - 3 } = \frac { z + 1 } { 1 }$ is
(1) 69
(2) 9
(3) $\sqrt { 69 }$
(4) $\sqrt { 99 }$