Notice that $ 3^{17} + 3^{10} = 3^{10} \cdot (3^7 + 1) $; thus $ 9 $ divides into $ 3^{17} + 3^{10} $. Furthermore, using the sum of seventh powers factorization, it follows that $ 3+1 = 4 $ divides into $ 3^7 + 1 $. Using the divisibility criterion for $ 4 $, we know that $ \overline{AB} $ must be divisible by $ 4 $. Thus $ B $ is even and not divisible by $ 3 $. Also, $ A $ is odd, so $ \overline{AB} = 10A + B $, where $ 4 $ does not divide into $ 10A $. Thus, $ 4 $ cannot divide into $ B $ either, otherwise $ 10A + B $ would not be divisible by $ 4 $. Then, $ B $ must be equal to $ 2 $. Using the divisibility criterion for $ 9 $, it follows that $ 3(A+B+C) $ is divisible by $ 9 $, that is $ 3 $ divides into $ A+C+2 $. Thus, $ A+C = 4,7,10,13,16 \quad (*) $. Using the divisibility criterion for $ 11 $, since \begin{align*}10^{8} \cdot A + 10^7 \cdot B + \cdots + B &\equiv (-1)^8 \cdot A + (-1)^7 \cdot B + \cdots + B \\ &\equiv A - B + \cdots + B \\ &\equiv -1 \pmod{11},\end{align*}then the alternating sum of digits, which works out to be $ B+C-A \equiv -1 \pmod{11} $. Thus, $ 2+C-A $ is either equal to $ 10 $ or $ -1 $, so $ A-C = 3,-8 $. In the former case when $ A-C = 3 $, summing with $ (*) $ yields that $ 2A \in \{7,10,13,16,19\} $, of which only $ A = 5 $ fit the problem conditions. This yields that $ C = 2 $. However, we know that $ B $ and $ C $ are distinct, so we can eliminate this possibility. Thus, $ A-C = -8 $, of which only $ C = 9, A = 1 $ works. The answer is $ \boxed{129} $.