12. As shown in the figure, $ABCD$ is a tetrahedron, $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, $CD=13$. Let $d$ be the distance between the midpoints of $AB$ and $CD$. Find the value of $d^{2}$.
12. By the median formula, we get $m_{s}^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}\right.$ $\left.-a^{2}\right)$, so $P C^{2}=\frac{1}{4}\left[2(A C)^{2}+2(B C)^{2}\right.$ $\left.-(A B)^{2}\right]=\frac{1009}{4}$.
$$
\begin{array}{l} 
P D^{2}\left.=\frac{1}{4}\left[2(A D)^{2}+2(B D)^{2}\right]-(A B)^{2}\right] \\
=\frac{425}{4} . \\
\therefore \quad P Q^{2}=\frac{1}{4}\left[2(P C)^{2}+2(P D)^{2}\right. \\
\left.-(C D)^{2}\right]=137 . \quad \text { (to be continued) }
\end{array}
$$
(to be continued)