smc

We need $$abc=2(ab+bc+ac)\text{ or }(a-2)bc=2a(b+c).$$ Since $a\le b$ and $a,b,c$ are all positive$,ac\le bc$. From the first equation we get $abc\le 6bc$. Thus $a\le 6$. From the second equation we see that $a>2$. Thus $a\in \{3,4,5,6\}$. If $a=3$ we need $bc=6(b+c)\Rightarrow (b-6)(c-6)=36$. We get five roots $\{(3,7,42),(3,8,24),(3,9,18),(3,10,15),(3,12,12)\}$. If $a=4$ we need $2bc=8(b+c)\Rightarrow bc=4(b+c)\Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20),(4,6,12),(4,8,8)\}$. If $a=5$ we need $3bc=10(b+c)\Rightarrow 9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100$. We get one root $\{(5,5,10)\}$. If $a=6$ we need $4bc=12(b+c)\Rightarrow bc=3(b+c)\Rightarrow (b-3)(c-3)=9$. We get one root $\{(6,6,6)\}$. Thus, there are $5+3+1+1=\boxed{10}$ solutions.