The numbers $\overline{abc}, \overline{bca}, \overline{cab}$ have the same sum of digits therefore they will have the same remainder $r$ when divided by 9.
Since the remainders modulo 27 are small, they are preserved modulo 9. Indeed if $n = 27k + r$, then $n = 9 \cdot 3k + r$, so $r$ will be a common remainder.
27 must divide the difference $\overline{bca} - \overline{abc} = 90b + 9c - 99a = 9(a + b + c) + 81b - 108a = 9(a + b + c) + 27(3b - 4a)$, therefore $3 \mid a+b+c$ which implies $r = 3$.

The numbers $abc$, with $a < b < c$ and $a+b+c = 12$ are 129, 138, 147, 156, 237, 246 and 345. Only 138 and 246 have the remainder 3 modulo 27.
The numbers $abc$, with $a < b < c$ and $a + b + c = 21$ are 489, 579 and 678. Convenient are 489 and 678.
In the end we have 4 solutions: 138, 246, 489 and 678.