Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let $P(-1,-1,1)$ and $Q(1,1,1)$. P lies on the plane $(-3-2+4+1=0)$, so its image is P itself. Let $Q^{\prime }$ be the image of Q and M the foot of the perpendicular from Q on the plane. The figure is triangle $PQ{Q}^{\prime }$, with area $2\cdot \frac{1}{2}\cdot QM\cdot PM.QM=\frac{|3+2+4+1|}{\sqrt{29}}=\frac{10}{\sqrt{29}}$ and $P{M}^2=P{Q}^2-Q{M}^2=8-\frac{100}{29}=\frac{132}{29}$. Area $=\frac{10}{\sqrt{29}}\cdot \frac{\sqrt{132}}{\sqrt{29}}=\frac{20\sqrt{33}}{29}$.