$$XYZ=2(X+Y+Z)$$ It is first noted that $X,Y,Z$ cannot all be the same, otherwise one would get $X^3=6X$ hence $X=\sqrt6$ Wlog, let $X \le Y \le Z$ we have $XYZ \le 6Z$ $\Rightarrow XY \le 6$ The only step left consists in checking all the possibilities satisfying the last equation (and there is no other way), namely $(X=1, Y=1), (X=1, Y=2), (X=1, Y=3), (X=2, Y=3), (X=1, Y=6), (X=2, Y=2)$. Only $(X=1, Y=3)$ and $(X=2, Y=2)$ do not yield fractionary, negative or absurd results for $Z$. Thus there are only two solutions, namely $(X=1, Y=3, Z=8)$ and $(X=2, Y=2, Z=4)$.

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In fact, I forgot to check not only $X=1, Y=4$, as pointed out by #3 (Thx), but also $X=1, Y=5$, providing the fourth solution $(1,5,4)$ But more importantly, we have to consider all the possible permutations of each triplet, for the sake of symmetry of the starting equation. i.e, if (1, 5, 4) is solution, then (4,5,1), (5, 1, 4), (1,4,5), (4,1,5) are also solutions. Sorry for having overlooked symmetry.