smc

Suppose the dice have $a$ and $b$ faces, and WLOG $a\ge b$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\frac{4}{3}\cdot 6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\ge 9$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\ge 9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\frac{n}{8a}=\frac{1}{12}⟹n=\frac{2}{3}a$. Since $b=8$, $n\le 8⟹a\le 12$. In addition to $3\mid a$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{17}$.