jmc

This approach uses analytical geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$, $C$ at $(20,0,20)$, and $D$ at $(20,20,20)$. Thus, $P$ is at $(5,0,0)$, $Q$ is at $(20,0,15)$, and $R$ is at $(20,10,20)$. Let the plane $PQR$ have the equation $ax+by+cz=d$. Using point $P$, we get that $5a=d$. Using point $Q$, we get $20a+15c=d⟹4d+15c=d⟹d=-5c$. Using point $R$, we get $20a+10b+20c=d⟹4d+10b-4d=d⟹d=10b$. Thus plane $PQR$’s equation reduces to $\frac{d}{5}x+\frac{d}{10}y-\frac{d}{5}z=d⟹2x+y-2z=10$. We know need to find the intersection of this plane with that of $z=0$, $z=20$, $x=0$, and $y=20$. After doing a little bit of algebra, the intersections are the lines $y=-2x+10$, $y=-2x+50$, $y=2z+10$, and $z=x+5$. Thus, there are three more vertices on the polygon, which are at $(0,10,0)(0,20,5)(15,20,20)$. We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are $5\sqrt{5}$. The other two are of $45-45-90△$s with legs of length 15, so their hypotenuses are $15\sqrt{2}$. So we have a hexagon with sides $15\sqrt{2},5\sqrt{5},5\sqrt{5},15\sqrt{2},5\sqrt{5},5\sqrt{5}$ By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it $20\sqrt{2}$. The height of the triangles at the top/bottom is $\frac{20\sqrt{2}-15\sqrt{2}}{2}=\frac{5}{2}\sqrt{2}$. The Pythagorean Theorem gives that half of the base of the triangles is $\frac{15}{\sqrt{2}}$. We find that the middle rectangle is actually a square, so the total area is $(15\sqrt{2}{)}^2+4\left(\frac{1}{2}\right)\left(\frac{5}{2}\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right)=\boxed{525}$.