Solution:
We will show that the only solutions are 135 and 144.
We have $a>0, b>0, c>0$ and
$$9(11 a+b)=(a+b+c)(a b c-1)$$
- If $a+b+c \equiv 0\pmod{3}$ and $a b c-1 \equiv 0\pmod{3}$, then $a \equiv b \equiv c \equiv 1\pmod{3}$ and $11 a+b \equiv 0\pmod{3}$. It follows now that
$$a+b+c \equiv 0\pmod{9} ; \text{ or } a b c-1 \equiv 0\pmod{9}$$
- If $a b c-1 \equiv 0\pmod{9}$
we have $11 a+b=(a+b+c) k$, where $k$ is an integer
and is easy to see that we must have $1<k \leq 10$.
So we must deal only with the cases $k=2,3,4,5,6,7,8,9,10$
If $k=2,4,5,6,8,9,10$ then $9 k+1$ has prime divisors greater than 9, so we must see only the cases $k=3,7$.
- If $k=3$ we have
$$8 a=2 b+3 c \text{ and } \quad a b c=28$$
It is clear that $c$ is even, $c=2 c_1$ and that both $b$ and $c_1$ are odd.
From $a b c_1=14$ it follows that $a=2, b=7, c_1=1$ or $a=2, b=1, c_1=7$ and it is clear that there exists no solution in this case.
- If $k=7$ we have $c$ even, $c=2 c_1$, $2 a=3 b+7 c_1$, $a b c_1=32$, then both $b$ and $c_1$ are even.
But this is impossible, because they imply that $a>9$.
Now we will deal with the case when $a+b+c \equiv 0\pmod{9}$ or $a+b+c=9 l$, 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 $a b c \geq 72$ and $a b c(a+b+c)>1000$, so the case $l \geq 2$ is impossible.
- If $l=1$ we have
$$11 a+b=a b c-1 \text{ or } 11 a+b+1=a b c \leq\left(\frac{a+b+c}{3}\right)^{3}=27$$
So we have only two cases: $a=1$ or $a=2$.
- If $a=1$, we have $b+c=8$ and $11+b=b c-1$ 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 $b(2 c-1)=23$ and there is no solution for the problem.