I am looking to solve the following equation in positive integers:
$$abc=5(a-2)(b-2)(c-2), a,b,c\in\mathbb{N}_0$$
From experimentation, I found that the only solution is $(10,4,4)$ and permutations, but I am not sure how to rigorously prove this. I simply looked at common prime factors and tried small values of the variables.
Let $a,b,c\in\Bbb{N}$ be such that $$abc=5(a-2)(b-2)(c-2).\tag{1}$$ By symmetry, without loss of generality we may assume that $a\leq b\leq c$.
If $abc=0$ then $a=0$, and so either $b=2$ or $c=2$. A quick check shows that $(a,b,c)$ is one of $$(0,0,2),\quad(0,1,2),\quad(0,2,c),\ c\geq2.$$ Now suppose $a,b,c>0$. Then the left hand side of $(1)$ is positive, hence the right hand side is positive, and so $a,b,c\geq3$. From $3\leq a\leq b\leq c$ we find that $$1=5(1-\tfrac2a)(1-\tfrac2b)(1-\tfrac2c)\geq5(1-\tfrac2a)^3,$$ and because $\left(\tfrac35\right)^3>\tfrac15$ this shows that $a<5$, so either $a=3$ or $a=4$.
If $a=3$ then, after some rearranging, equation $(1)$ simplifies to $$(b-5)(c-5)=15.$$ Because $3\leq b\leq c$ we see that $(a,b,c)$ is either $(3,6,20)$ or $(3,8,10)$.
If $a=4$ then, after some rearranging, equation $(1)$ simplifies to $$(3b-10)(3c-10)=40.$$ Because $4\leq b\leq c$ we see that $(a,b,c)$ is either $(4,4,10)$ or $(4,5,6)$.