Square pyramid $ABCDE$ has base $ABCD$ with side length $3$ cm and altitude $\overline{AE}$ perpendicular to the base with $AE=12$ cm. Point $P$ lies on $\overline{BE}$, one third of the way from $B$ to $E$; point $Q$ lies on $\overline{DE}$, one third of the way from $D$ to $E$; point $R$ lies on $\overline{CE}$, halfway from $C$ to $E$. What is the area of $\triangle PQR$, in square centimeters?

1. Use $A=(0,0,0)$, $B=(3,0,0)$, $C=(3,3,0)$, $D=(0,3,0)$, $E=(0,0,12)$. Then $P=B+\frac13(E-B)=(2,0,4)$, $Q=D+\frac13(E-D)=(0,2,4)$, $R=\frac{C+E}2=(\frac32,\frac32,6)$.
2. $PQ=\sqrt{4+4}=2\sqrt2$, and $PR=\sqrt{\frac14+\frac94+4}=\sqrt{\frac{13}2}=QR$.
3. So $\triangle PQR$ is isosceles with base $2\sqrt2$ and legs $\sqrt{13/2}$; its height is $\sqrt{\frac{13}2-2}=\frac3{\sqrt2}$.
4. Area $=\frac12\cdot2\sqrt2\cdot\frac3{\sqrt2}=\boxed{3}$.