Wait, what?

$\begin{aligned} \overline{aaaaaa} & = b \cdot c \cdot \overline{aa} \cdot \overline{ab} \cdot \overline{bc} \\ 111111a & = bc(11a)(10a+b)(10b+c) & \small \color{#3D99F6} \text{Divide both sides by }11a. \\ 10101 & = bc(10a+b)(10b+c) \\ {\color{#3D99F6}3} \cdot {\color{#D61F06}7} \cdot 1{\color{#3D99F6}3} \cdot {\color{#3D99F6}3}{\color{#D61F06}7} & = {\color{#3D99F6}b} \cdot {\color{#D61F06}c} \cdot \overline{a{\color{#3D99F6}b}} \cdot \overline{{\color{#3D99F6}b}{\color{#D61F06}c}} \end{aligned}$

Since there is only one 1 in the LHS and one $a$ in the RHS, $a=1$ , then $\overline{ab} = 13$ , $\implies b = 3$ and $c=7$ . $\implies a+b+c = 1+3+7 = \boxed{11}$ .