2. As shown in Figure 1, each face of the cube has a positive integer. It is known that the sum of the numbers on opposite faces is equal. If the numbers opposite to $13, 9, 3$ are $a, b, c$ respectively, then the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is ( ).
(A) 48
(B) 76
(C) 96
(D) 152
2. B.

According to the problem, we have
$$
13+a=9+b=3+c \text {. }
$$

Thus, $a-b=-4, b-c=-6, c-a=10$.
Then $a^{2}+b^{2}+c^{2}-a b-b c-c a$
$$
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=76 .
$$