jmc

Adding the three congruences gives $$\begin{array}{rl}& 6(a+b+c)\equiv 8(\mathrm{mod}7)\\ ⟹& -(a+b+c)\equiv 1(\mathrm{mod}7).\end{array}$$Adding this to each of the congruences yields $$\begin{array}{rl}b+2c& \equiv 1(\mathrm{mod}7),\\ a+2b& \equiv 5(\mathrm{mod}7),\\ 2a+c& \equiv 5(\mathrm{mod}7).\end{array}$$Substituting $b\equiv 1-2c(\mathrm{mod}7)$ into the second one gives $$\begin{array}{rl}& a+2(1-2c)\equiv 5(\mathrm{mod}7)\\ ⟹& a-4c\equiv 3(\mathrm{mod}7)\\ ⟹& 4c-a\equiv 4(\mathrm{mod}7)\\ ⟹& 8c-2a\equiv 8(\mathrm{mod}7)\\ ⟹& c-2a\equiv 1(\mathrm{mod}7).\end{array}$$Adding this to $2a+c\equiv 5(\mathrm{mod}7)$ results in $2c\equiv 6(\mathrm{mod}7)⟹c\equiv 3(\mathrm{mod}7)$. Finally $$\begin{array}{rl}& b\equiv 1-2c\equiv 1-2\cdot 3\equiv 2(\mathrm{mod}7),\\ & a\equiv 5-2b\equiv 5-2\cdot 2\equiv 1(\mathrm{mod}7).\end{array}$$Thus, $abc\equiv 1\cdot 2\cdot 3\equiv \boxed{6}$.