jmc

Dealing with the second condition, it is set up in the following manner. \begin{array}{c@{}c@{}c@{}c} &&A&B_6\\ &+&&C_6\\ \cline{2-4} &&C&0_6\\ \end{array}Because $B+C$ can not equal $0$, we must carry in this column. Therefore, we arrive at two equations. $B+C-6=0$ and $A+1=C$ Looking at the third condition: \begin{array}{c@{}c@{}c@{}c} &&A&B_6\\ &+&B&A_6\\ \cline{2-4} &&C&C_6\\ \end{array}We can determine that no carrying occurs. Therefore, $A+B=C$. We now have a system of equations for our three variables. $$B+C-6=0$$$$A+1=C$$$$A+B=C$$Subtracting the third equation from the second, $1-B=0$, or $B=1$. Plugging that into our first equation we can determine that $C=5$. $A$ must then equal $4$. Thus, $\overline{ABC}$ is $\boxed{415}$.