1. Given $y z \neq 0$, and the set $\{2 x, 3 z, x y\}$ can also be represented as $\left\{y, 2 x^{2}, 3 x z\right\}$. Then $x=$ $\qquad$
,- 1.1 .
It is clear that $x y z \neq 0$.
From the equality of the product of elements of the two sets, we get
$$
6 x^{2} y z=6 x^{3} y z \Rightarrow x=1 \text{. }
$$

Upon verification, $x=1$ satisfies the problem conditions.