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

The correct option is C $x-4y+7z=0$


Let image of point $(3,1,7)$ with respect to the plane $x-y+z=3$ is $P(h,k,l)$
$\therefore \frac{h-3}{1}=\frac{k-1}{-1}=\frac{l-7}{1}=\frac{-2(3-1+7-3)}{3}=-4$
After solving, we get
$h=-1,k=5,l=3$
Therefore, the image is $P(-1,5,3)$

Also, required plane contains the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$
$\Rightarrow$ The plane passes through $(0,0,0)$ and $(1,2,1)$
Therefore, required equation of the plane passing through the three points is
$\left|\begin{array}{ccc}x-0& y-0& z-0\\ 0-1& 0-2& 0-1\\ 0-(-1)& 0-5& 0-3\end{array}\right|=0$

$\Rightarrow x-4y+7z=0$