, 2017

Solution:
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 $1+b c=84 \Longrightarrow b c=83$. 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 $(a b+1)(b c+1)(c a+1) \geq 5^3 > 84$. So WLOG, set $a=1$. We now need $(b+1)(c+1)(b c+1)=84$. Now, WLOG, say $b \leq c$. If $b=1$, then $(c+1)^2=42$, which has no solution. If $b \geq 3$, then $(b+1)(c+1)(b c+1) \geq 4^2 \cdot 10 = 160 > 84$. 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 $6+6=12$.