, 2019

Solution:

Let $A D$ be the positive $x$-axis, $A B$ be the positive $y$-axis, and $A E$ be the positive $z$-axis, with $A$ the origin. The plane, which passes through the origin, has equation $k_{1} x + k_{2} y = z$ for some undetermined parameters $k_{1}, k_{2}$. Because $A P = A S$ and $A B = A D$, we get $P B = S D$, so $P$ and $S$ have the same $z$-coordinate. But $P(0, 6, 6 k_{2})$ and $S(6, 0, 6 k_{1})$, so $k_{1} = k_{2} = k$ for some $k$. Then $Q$ and $R$ both have $z$-coordinate $49$, so $Q\left(\frac{49}{k} - 6, 6, 49\right)$ and $R\left(6, \frac{49}{k} - 6, 49\right)$. The equation $Q R^{2} = R S^{2}$ then gives
$$\left(\frac{49}{k} - 6\right)^{2} + (49 - 6k)^{2} = 2\left(12 - \frac{49}{k} - 12\right)^{2}$$
This is equivalent to
$$(49 - 6k)^{2}(k^{2} + 1) = 2(49 - 12k)^{2}$$
which factors as
$$(k - 7)(36k^{3} - 336k^{2} - 203k + 343) = 0.$$
This gives $k = 7$ as a root. Note that for $Q$ and $R$ to actually lie on $F G$ and $G H$ respectively, we must have $\frac{49}{6} \geq k \geq \frac{49}{12}$. Via some estimation, one can show that the cubic factor has no roots in this range (for example, it's easy to see that when $k = 1$ and $k = \frac{336}{36} = \frac{28}{3}$, the cubic is negative, and it also remains negative between the two values), so we must have $k = 7$.

Now consider projecting $A P Q R S$ onto plane $A B C D$. The projection is $A B C D$ save for a triangle $Q' C R'$ with side length $12 - \frac{49}{k} = 5$. Thus the projection has area $36 - \frac{25}{2} = \frac{47}{2}$. Since the area of the projection equals $[A P Q R S] \cdot \cos \theta$, where $\theta$ is the (smaller) angle between planes $A P Q R S$ and $A B C D$, and since the planes have normal vectors $(k, k, -1)$ and $(0, 0, 1)$ respectively, we get $\cos \theta = \frac{(k, k, -1) \cdot (0, 0, 1)}{\sqrt{k^{2} + k^{2} + 1}} = \frac{1}{\sqrt{2k^{2} + 1}} = \frac{1}{\sqrt{99}}$ and so
$$[A P Q R S] = \frac{47 \sqrt{99}}{2} = \frac{141 \sqrt{11}}{2}$$