1. Given three positive integers $x, y, z$ whose least common multiple is 300, and $\left\{\begin{array}{l}x+3 y-2 z=0, \\ 2 x^{2}-3 y^{2}+z^{2}=0\end{array}\right.$. Then the solution to the system of equations $(x, y, z)=$ $\qquad$ .
2. $(20,60,100)$.
Transform $x+3 y-2 z=0$ and substitute into $2 x^{2}-3 y^{2}+z^{2}=0$, eliminating $x$ we get
$$
5 y^{2}-8 y z+3 z^{2}=0 \text {, }
$$

which is $(5 y-3 z)(y-z)=0$.
Thus, $5 y-3 z=0$,

or $y-z=0$.
From equation (1) and $x+3 y-2 z=0$, we get $y=3 x, z=5 x$, i.e., $x: y: z=1: 3: 5$.
Therefore, by the given conditions, we must have
$$
x=20, y=60, z=100 \text {. }
$$

From equation (2) and $x+3 y-2 z=0$, we get $x=-y=-z$, which contradicts the given conditions.