[tex]\overline{a,b}+\overline{b,c}+\overline{a,c}=\frac{77}{5}\\
\frac{\overline{ab}}{10}+\frac{\overline{bc}}{10}+\frac{\overline{ac}}{10}=\frac{77}{5}|*10\\
\overline{ab}+\overline{bc}+\overline{ac}=154\\
10a+b+10b+c+10a+c=154\\
11a+11b+11c=154\\
11(a+b+c)=154\\
\boxed{a+b+c=14}[/tex]