smc

Using the given data, we can label the points $A(0,0,0),B(3,0,0),C(3,3,0),D(0,3,0),$ and $E(0,0,6)$. We can also find the points $P=B+\frac{1}{3}\overrightarrow{BE}=(3,0,0)+\frac{1}{3}(-3,0,6)=(3,0,0)+(-1,0,2)=(2,0,2)$. Similarly, $Q=(0,2,2)$ and $R=(1,1,4)$. Using the distance formula, $PQ=\sqrt{{\left(-2\right)}^2+2^2+0^2}=2\sqrt{2}$, $PR=\sqrt{{\left(-1\right)}^2+1^2+2^2}=\sqrt{6}$, and $QR=\sqrt{1^2+{\left(-1\right)}^2+2^2}=\sqrt{6}$. Using Heron's formula, or by dropping an altitude from $P$ to find the height, we can then find that the area of $△PQR$ is $\boxed{2\sqrt{2}}$. Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\overrightarrow{PQ}=[-2,2,0]$ and $\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\text{area}=\frac{1}{2}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\sqrt{2}$, giving the area as $2\sqrt{2}$ like before.