Restricted Partition

To make this more simpler, let's substitute,
$a = 11 + l$
$b = 21 + m$
$c = 31 + n$

Thus, the equality becomes,
$(l+m+n)+63 = 177$
$\Rightarrow l+m+n = 114$

Now, we know that the number of integer solutions to,
$x_1+x_2+\ldots+x_r = n$ ,
where $x_1,x_2,\ldots,x_r \geq 0$ are ${n+r-1}\choose{r-1}$

Thus, the number of solutions to our equation,
$l+m+n = 114$
would be,

${114+3-1}\choose{3-1}$ $=$ $116\choose 2$ $= 6670$

Note that the last three digits of $6670$ are $\boxed{670}$

same like anish