We will show that the only solutions are 135 and 144.

We have $a > 0, b > 0, c > 0$ and

$$9(11a + b) = (a + b + c)(abc - 1)$$

- If $a + b + c \equiv 0 \pmod{3}$ and $abc - 1 \equiv 0 \pmod{3}$, then $a \equiv b \equiv c \equiv 1 \pmod{3}$ and $11a + b \equiv 0 \pmod{3}$. It follows now that

$$a + b + c \equiv 0 \pmod{9} \text{ or } abc - 1 \equiv 0 \pmod{9}$$

- If $abc - 1 \equiv 0 \pmod{9}$

we have $11a + b = (a + b + c)k$, where $k$ is an integer

and it is easy to see that we must have $19$.

Now we will deal with the case when $a + b + c \equiv 0 \pmod{9}$ or $a + b + c = 9l$, where $l$ is an integer.

- If $l \geq 2$ we have $a + b + c \geq 18, \max\{a, b, c\} \geq 6$ and it is easy to see that $abc \geq 72$ and $abc(a + b + c) > 1000$, so the case $l \geq 2$ is impossible.
- If $l = 1$ we have

$$11a + b = abc - 1 \text{ or } 11a + b + 1 = abc \leq \left(\frac{a + b + c}{3}\right)^3 = 27$$

So we have only two cases: $a = 1$ or $a = 2$.

- If $a = 1$, we have $b + c = 8$ and $11 + b = bc - 1$ or $b + (c - 1) = 7$ and $b(c - 1) = 12$ and the solutions are $(a, b, c) = (1, 3, 5)$ and $(a, b, c) = (1, 4, 4)$, and the answer is 135 and 144.
- If $a = 2$ we have $b(2c - 1) = 23$ and there is no solution for the problem.