jmc

By definition of a base, it follows that ${\overline{ABC}}_6=6^2\cdot A+6\cdot B+C$. Noting that every digit appears in every possible slot once, it follows that ${\overline{ABC}}_6+{\overline{BCA}}_6+{\overline{CAB}}_6=(6^2+6+1)(A+B+C).$ The value is equal to the sum, ${\overline{AAA0}}_6=6^3\cdot A+6^2\cdot A+6\cdot A=(6^2+6+1)\cdot (6\cdot A)$. Setting them equal, $$(6^2+6+1)(A+B+C)=(6^2+6+1)\cdot (6\cdot A)⟹B+C=5\cdot A.$$Since $B,C<6$, then $B+C<2\cdot 6$, so $A=1,2$. However, there do not exist distinct base $6$ digits such that $B+C=2\cdot 5$, so it follows that $A=1_6$, and $B+C={\boxed{5}}_6$.