Problem 961

We place the points in the coordinate plane. We let $A=\left(0,0, \frac{\sqrt{6}}{3}\right), B=\left(0, \frac{\sqrt{3}}{3}, 0\right)$, $C=\left(-\frac{1}{2},-\frac{\sqrt{3}}{6}, 0\right)$, and $D=\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, 0\right)$. The point $P$ is the origin, while $M$ is $\left(0,0, \frac{\sqrt{6}}{6}\right)$. The line through $B$ and $M$ is the line $x=0, y=\frac{\sqrt{3}}{3}-z \sqrt{2}$. The plane through $A, C$, and $D$ has equation $z=2 \sqrt{2} y+\sqrt{\frac{2}{3}}$. The coordinates of $Q$ are the coordinates of the intersection of this line and this plane. Equating the equations and solving for $y$ and $z$, we see that $y=-\frac{1}{5 \sqrt{3}}$ and $z=\frac{\sqrt{6}}{5}$, so the coordinates of $Q$ are $\left(0,-\frac{1}{5 \sqrt{3}}, \frac{\sqrt{6}}{5}\right)$. Let $N$ be the midpoint of $CD$, which has coordinates $\left(0,-\frac{\sqrt{3}}{6}, 0\right)$. By the distance formula, $QN=\frac{3 \sqrt{3}}{10}$. Thus, the area of $QCD$ is $\frac{QN \cdot CD}{2}=\frac{3 \sqrt{3}}{20}$.