The solutions are $(0,1,83)$ and $(1,2,3)$ up to permutation. First, we do the case where at least one of $a, b, c$ is 0. WLOG, say $a=0$. Then we have $1+bc=84 \Longrightarrow bc=83$. As 83 is prime, the only solution is $(0,1,83)$ up to permutation. Otherwise, we claim that at least one of $a, b, c$ is equal to 1. Otherwise, all are at least 2, so $(1+ab)(1+bc)(1+ac) \geq 5^{3}>84$. So WLOG, set $a=1$. We now need $(b+1)(c+1)(bc+1)=84$. Now, WLOG, say $b \leq c$. If $b=1$, then $(c+1)^{2}=42$, which has no solution. If $b \geq 3$, then $(b+1)(c+1)(bc+1) \geq 4^{2} \cdot 10=160>84$. So we need $b=2$. Then we need $(c+1)(2c+1)=21$. Solving this gives $c=3$, for the solution $(1,2,3)$. Therefore, the answer is $6+6=12$.