12.4.9 ** In a sphere with radius 1, there is an inscribed regular dodecahedron (the regular dodecahedron is symmetric with respect to the center of the sphere). Find the sum of the squares of the distances between all vertices of this regular dodecahedron.
Let $P, Q$ be two vertices of a regular dodecahedron, and $P^{\prime}, Q^{\prime}$ be their respective symmetric vertices, then $P P^{\prime}=2, Q Q^{\prime}=2$. And
$$
P Q^{2}+P^{\prime} Q^{2}-P P^{\prime 2}-4, P^{\prime} Q^{2}+P^{\prime} Q^{\prime 2}=Q Q^{\prime 2}=4 .
$$

Since a regular dodecahedron has 20 vertices, the sum of the squares of the distances between symmetric vertices is 40. Since the number of pairs of non-symmetric vertices $(P, Q)$ is $\mathrm{C}_{20}^{2}-10$, the sum of the squares of the distances between non-symmetric vertices is $\sum P Q^{2}=$ $2\left(\mathrm{C}_{20}^{2}-10\right)=360$. Therefore, the sum of the squares of the distances between all vertices of the regular dodecahedron is $360+$ $40=400$.