By taking their product we get: $$abc \vert (ab+1)(bc+1)(ac+1) \Rightarrow abc \vert ab+bc+ca+1$$$$\Rightarrow \frac{1}{abc}+\sum_{cyc} \frac{1}{a}=k \in \mathbb{N}$$WLOG $a \leq b \leq c$. $$a>3 \Rightarrow k \leq \frac{1}{64}+\frac{3}{4}<1$$Giving a contradiction so $a \leq 3$. If $a=3$ then if $b>3$ we get a contradiction in a similar way so $b=3$. Hence we get: $$k=\frac{10+6c}{9c} \Rightarrow 3 \vert 10+6c$$Which is clearly a contradiction. If $a=2$ then if $b>4$ we get: $$k \leq \frac{1}{50}+\frac{2}{5}+\frac{1}{2}<1$$Giving a contradiction so $2 \leq b \leq 4$. If $b=2$ we get: $$k=1+\frac{5}{4c} \Rightarrow 4 \vert 5$$ If $b=3$: $$k=\frac{5c+7}{6c} \Rightarrow 6c \vert 5c+7$$As $12c>5c+7$ for $c>1$ we must have: $$6c=5c+7 \Rightarrow c=7$$ If $b=4$: $$k=\frac{9+6c}{8c} \Rightarrow 8c \vert 9+6c$$But for $c>1$ we have $16c>9+6c$ hence we need: $$8c=9+6c \Rightarrow 2c=9$$Which is an obvious contradiction. So the only triplet it $(2,3,7)$ and symettric permutations which can indeed be checked to work.