jmc

Since $13$ is a prime, each of $a,b,c$ is invertible modulo $13$. Let $a^{-1}=x,b^{-1}=y,c^{-1}=z$ in modulo $13$. Multiplying both sides of each congruence by $(abc{)}^{-1}$ yields $$\begin{array}{rl}2z+x+y& \equiv 0(\mathrm{mod}13),\\ z+2x+y& \equiv 6(\mathrm{mod}13),\\ z+x+2y& \equiv 8(\mathrm{mod}13).\end{array}$$Adding all three together gives $4(x+y+z)\equiv 14(\mathrm{mod}13)⟹x+y+z\equiv 10(\mathrm{mod}13)$. Subtracting this from each results in $$\begin{array}{rl}z\equiv -10\equiv 3& (\mathrm{mod}13),\\ x\equiv -4\equiv 9& (\mathrm{mod}13),\\ y\equiv -2\equiv 11& (\mathrm{mod}13).\end{array}$$Thus, $a+b+c\equiv x^{-1}+y^{-1}+z^{-1}\equiv 9+3+6\equiv 18\equiv \boxed{5}(\mathrm{mod}13)$.