, 2015

Solution:
We can write $a + a b + a b c = a(1 + b + b c)$. Since $11$ is prime, $a = 11$ or $a = 1$. But since $b, c$ are both positive integers, we cannot have $a = 11$, and so $a = 1$.

Then $1 + b + b c = 11 \Longrightarrow b + b c = 10 \Longrightarrow b(c + 1) = 10$, and since $c$ is a positive integer, only $b = 1, 2, 5$ are possible. This gives the $3$ triples $(a, b, c) = (1, 1, 9), (1, 2, 4), (1, 5, 1)$.