After some calculations , we need to find non-negative solutions to the equation $abc+2=ab+bc+ca$(symmetric => if a solution set is valid, then all its permutations are as well) From here, $a=\frac{bc-2}{bc-b-c}=\frac{bc-b-c+b+c-2}{bc-b-c}=1+\frac{b+c-2}{bc-b-c}$ so $\frac{b+c-2}{bc-b-c}$ must be an integer, which further implies that $b+c-2 \geq bc-b-c $ $=>$ $2 \geq (b-2)(c-2)$ Thus $4 \geq b,c$ 1st Case: $c=1$ This implies $a+b=2$ thus $(a,b,c)=(0,2,1),(2,0,1),(1,1,1)$ 2nd Case: $c=2$ we have $ab-a-b=a+b-2$ , so $(a-2)(b-2)=2$ and this gives $(a,b,c)=(3,4,2),(4,3,2),(1,0,2),(0,1,2)$ 3rd Case:$c=3$ the equation becomes $2ab-3a-3b+2=0$, which is easy to verify $(a,b,c)=(4,2,3),(2,4,3)$ 4th Case: $c=4$ We have $3ab-4a-4b+2=0$ and agin by checking we get $(a,b,c)=(3,2,4),(2,3,4)$ Finally if $c=0$ ,$ab=2$ , thus $(a,b,c)=(2,1,0),(1,2,0)$