JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Three Dimensional Geometry

The image of point $(3,1,7)$ with respect to the plane $x-y+z=3$ is found using the formula for reflection. The perpendicular from $(3,1,7)$ to the plane has direction ratios $(1,-1,1)$. The foot of the perpendicular satisfies $\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=\frac{-2(3-1+7-3)}{3}=-4$. So the image is $P=(-1,5,3)$. The plane through $P$ and the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ has normal vector $\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 5& 3\\ 1& 2& 1\end{array}\right|=\hat{i}(5-6)-\hat{j}(-1-3)+\hat{k}(-2-5)=-\hat{i}+4\hat{j}-7\hat{k}$. The plane equation is $x-4y+7z=0$.