One, (Full marks 10 points) As shown in the figure, there is a natural number on each face of the cube. It is known that the sum of the two numbers on opposite faces is equal. Given that the numbers opposite to $12$, $9$, and $3$ are $a$, $b$, and $c$ respectively, find the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$.
Final 1. From the problem, we have $13+a=9+b=3+c$, which gives $a-b=-4, b-c=-6, c-a=10$. Therefore,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\
=\frac{1}{2}(16+36+100)=76 .
\end{array}
$$