Completeley straightforward. We need $abc \mid (ab-1)(bc-1)(ca-1)$. Since $\gcd(a, (ab-1)(ca-1)) = 1$, it follows $a\mid bc-1$, and similarly $b\mid ca-1$ and $c\mid ab -1$. These force $a,b,c$ to be pairwise coprime, so $abc \mid ab+bc+ca-1$. Then $abc \leq ab+bc+ca - 1$, so $1/a + 1/b + 1/c \geq 1 + 1/abc$. Assume $1\leq a \leq b \leq c$. If $a\geq 3$ then $1/a + 1/b + 1/c \leq 1/3 + 1/4 + 1/5 < 1 < 1 + 1/abc$. If $a= 2$ then $1/b + 1/c \geq 1/2 + 1/2bc$, so $2(b+c) \geq bc+1$. This writes $(b-2)(c-2) \leq 3$; but $b>2$ (since $a,b$ coprime), whence the unique possibility $b=3$, $c=5$ ($c=3$ and $c=4$ are ineligible because of coprimality). If $a= 1$ then $1/b + 1/c \geq 1/bc$ for all $b,c$. We then need $bc \mid b+bc+c-1$, so $bc \mid b+c-1$, therefore $bc \leq b+c-1$. This writes $(b-1)(c-1) \leq 0$, forcing $b=1$, and we can see any $c\geq 1$ will do. Thus the solutions are $\{a,b,c\} = \{2,3,5\}$ (six solutions), and the infinite families $(1,1,n), (1,n,1), (n,1,1)$ for all $n\geq 1$.