7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$.
7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:

$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$

This inequality becomes an equality if the numbers are equal, i.e., $2 x=3 x y^{2}=2 z=12 \Rightarrow z=6, x=6, y=\sqrt{2 / 3}$. Therefore, the maximum value is 144.

Answer: $\{144\}$