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

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
$ax+by+cz+d=0$
it contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$
so $d=0$
and $a+2b+c=0$
also $-a+5b+3c=0$
$\therefore \frac{a}{1}=\frac{b}{-4}=\frac{c}{7}$
$\therefore$ equation of plane is
$x-4y+7z=0$