Assume $b$ and $c$ exist. Considering the two values modulo 13, we find $$\begin{cases}a+10 b+2 c \equiv 2 & (\bmod 13) \\ 2 a+5 b+c \equiv 8 & (\bmod 13)\end{cases}$$ Subtracting twice the second equation from the first, we get $-3 a \equiv-14(\bmod 13)$. So, we have $a \equiv 9$ $(\bmod 13)$. Therefore we must either have $a=9$ or $a=22$. Moreover, both $a=9$ and $a=22$ yield solutions with $b=0$ and $c=3,16$, depending on the value of a. Thus the answer is $9+22=31$.