JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Three Dimensional Geometry

The reflection of $P(1,2,1)$ in the plane $x+2y+2z=16$ uses the normal $\vec{n}=(1,2,2)$ with $|\vec{n}{|}^2=9$. Since $x+2y+2z-16=-9$ at $P$, the image is $Q=P-\frac{2(-9)}{9}\vec{n}=P+2\vec{n}=(3,6,5)$. The plane $T$ contains the given line, which passes through $(0,0,-1)$ with direction $(1,1,2)$, and also passes through $Q$. Taking $\overrightarrow{AQ}=(3,6,6)$, a normal to $T$ is $(1,1,2)\times (3,6,6)=(-6,0,3)\parallel (2,0,-1)$, so $T:2x-z=1$. Of the given points, only $(1,2,1)$ satisfies $2x-z=1$.