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{align*} 2z+x+y&\equiv 0 \pmod{13},\\ z+2x+y&\equiv 6 \pmod{13},\\ z+x+2y&\equiv 8 \pmod {13}.\end{align*} Adding all three together gives $ 4(x+y+z)\equiv 14\pmod {13}\implies x+y+z\equiv 10\pmod {13} $. Subtracting this from each results in \begin{align*} z\equiv -10\equiv 3&\pmod{13},\\ x\equiv -4\equiv 9&\pmod{13},\\ y\equiv -2\equiv 11&\pmod {13}.\end{align*} Thus, $ a+b+c\equiv x^{-1}+y^{-1}+z^{-1}\equiv 9+3+6\equiv 18\equiv \boxed{5}\pmod{13} $.