This is equivalent to the positive integer solutions to $$\frac{a-2}{a} + \frac{b-2}{b} + \frac{c-2}{c} = 2$$ with $3 \le a \le b \le c$.
Small solutions like $(6, 6, 6)$ and $(4, 8, 8)$ can be guessed, but other solutions such as $(4, 5, 20)$ exist.
Well, basically ${1\over a}+{1\over b} +{1\over c}={1\over2}$ is what you want.
Let $a\geq b\geq c$ we know since ${({1\over2})\over3}={1\over6}$, ${1\over c}\geq{1\over6}$ and hence $c\leq 6$.
This leaves us four possibilities.
(1) $c=3$ then ${1\over a}+{1\over b}={1\over 6}\implies ab=6a+6b\implies (a-6)(b-6)=36$
(2) $c=4$ then ${1\over a}+{1\over b}={1\over 4}\implies ab=4a+4b\implies (a-4)(b-4)=16$
(3) $c=5$ then ${1\over a}+{1\over b}={3\over 10}\implies 3ab=10a+10b\implies (3a-10)(3b-10)=100$
(4) $c=6$ then ${1\over a}+{1\over b}={1\over 3}\implies ab=3a+3b\implies (a-3)(b-3)=9$