jmc

We start off by replacing $C$ with $A+B$ and changing the form of the two-digit integer in the second equation. $$\begin{array}{rl}10A+A-B& =2\times (A+B)\Rightarrow \\ 11A-B& =2A+2B\Rightarrow \\ 9A& =3B\Rightarrow \\ 3A& =B\end{array}$$Now we replace $C$, change the two-digit integer, and then substitute $B$ with $3A$ in the third equation. $$\begin{array}{rl}(A+B)\times B& =10A+A+A\Rightarrow \\ & =12A\Rightarrow \\ (A+3A)\times 3A& =12A\Rightarrow \\ (4A)\times 3A& =12A\Rightarrow \\ 12(A{)}^2& =12A\end{array}$$For $(A{)}^2$ to equal $A$, $A$ must equal 1. Since $3A=B$, $B=3$. That means $A+B=C=4$. So the sum of the three digits is $1+3+4=\boxed{8}$.