A. B. C

We are given that $\overline{AA} + \overline{BB} + \overline{CC} = \overline{ABC} \Longrightarrow$

$10A + A + 10B + B + 10C + C = 100A + 10B + C \Longrightarrow B + 10C = 89A$ .

Now since $A,B,C$ are digits the maximum of $B + 10C$ is $99$ , thus we must have $A = 1$ . We then need to find $B,C$ such that $B + 10C = 89$ . Since $B \le 9$ we can only have $C = 8$ , and thus $B = 9$ .

The unique value for $\overline{ABC}$ is therefore $\boxed{198}$ .

AA+BB+CC=ABC or 11+99+88=198 this problem can be solved by logic A=1 B=9 C=8 then AA=11, BB=99,CC=88 so, ABC=198 BY:YNT