10. All positive integer solutions to the equation $5(x y+y z+z x)=4 x y z$, with $x \leqslant y \leqslant z$ are
$\qquad$ .
10. $(2,5,10),(2,4,20)$

Original equation $\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$.
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x}$, we get $1<x \leqslant \frac{15}{4}$, so $x=2$ or 3, and we can discuss the following cases accordingly.