JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Three Dimensional Geometry

The plane is the perpendicular bisector of the segment joining $(1,2,3)$ to its mirror image $\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$. The normal direction is the difference of the two points, $\left(-\frac{10}{3},-\frac{10}{3},-\frac{10}{3}\right)$, which is parallel to $(1,1,1)$, and the plane passes through their midpoint $\left(-\frac{2}{3},\frac{1}{3},\frac{4}{3}\right)$. Putting the midpoint into $x+y+z=d$ gives $d=-\frac{2}{3}+\frac{1}{3}+\frac{4}{3}=1$, so the plane is $x+y+z=1$. Checking the offered points, only $(1,-1,1)$ gives $1-1+1=1$ and so lies on this plane.