, 2021

Solution:

We can divide the cube into 3 regions based on the value of $x+y+z$ which defines the plane: $x+y+z<1$, $1 \leq x+y+z \leq 2$, and $x+y+z>2$. The two regions on the ends create tetrahedra, each of which has volume $1/6$. The middle region is a triangular antiprism with volume $2/3$.

If our point $P$ lies in the middle region, we can see that we will always get the same value $3\sqrt{2}$ for the perimeter of $\mathcal{R}$.

Now let us compute the expected perimeter given that we pick a point $P$ in the first region $x+y+z<1$. If $x+y+z=a$, then the perimeter of $\mathcal{R}$ will just be $3\sqrt{2}a$, so it is sufficient to find the expected value of $a$. $a$ is bounded between $0$ and $1$, and forms a continuous probability distribution with value proportional to $a^2$, so we can see with a bit of calculus that its expected value is $3/4$.

The region $x+y+z>2$ is identical to the region $x+y+z<1$, so we get the same expected perimeter. Thus we have a $2/3$ of a guaranteed $3\sqrt{2}$ perimeter, and a $1/3$ of having an expected $\frac{9}{4}\sqrt{2}$ perimeter, which gives an expected perimeter of $\frac{2}{3} \cdot 3\sqrt{2} + \frac{1}{3} \cdot \frac{9}{4}\sqrt{2} = \frac{11\sqrt{2}}{4}$.

The square of this is $\frac{121}{8}$, giving an extraction of $12108$.