@Gyrate’s spoiler
Since the numbers are AA, BB, and CC we could rewrite as 10A + A, 10B + B, and 10C + C. That’s 11(A+B+C) so you know ABC has to be a multiple of 11. Considering that it’s over 100, the even multiples 121, 132, 143, and so on show that B-A=C. As you say, AA + BB =110 so the last digit will be C. And as you say the number can’t be over 300, since 99 + 99 + 99 is only 297. If we try the 2 as A and 8 as B, then C would be 6. But 22 + 88 + 66 isn’t 286. That leaves A=1, so B must be 9, and C must be 8. The other solution: -1, -9, -8.