Question

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{ x }{1}=\frac{ y }{2}=\frac{ z }{1}$ is

• A. $x+y-3 z=0$
• B. $3 x + z =0$
• C. $x-4 y+7 z=0$
• D. $2 x-y=0$

Answer 1

Answer: (C)

Mirror image of $(3,1,7)$ w.r.t $x-y+z=3$ is given by
$\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\frac{-2(3-1+7-3)}{3} $
$x--1, y-5, z-3 $
$P(-1,5,3)$
Let equation of the required plane is
$a x+b y+c z+d=0$
it contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$
so $d =0$
and $a+2 b+c=0$
also $-a+5 b+3 c=0$
$\therefore \frac{ a }{1}=\frac{ b }{-4}=\frac{ c }{7}$
$\therefore $ equation of plane is
$x-4 y+7 z=0$