In the given equation:
$$216^{2/3}÷{\left(\frac{p^2}{9}\right)}^{-3/2}=2^x{3}^y{p}^z$$

What are the values of $x$, $y$, and $z$?

Let us simplify the left-hand side (LHS) of the equation step-by-step:
1. Express 216 in prime base forms:

$$216=6^3=(2\times 3{)}^3=2^3\times 3^3$$

Therefore, the first term becomes:

$$216^{2/3}={\left(6^3\right)}^{2/3}=6^2=2^2\times 3^2$$

2. Simplify the second term with negative fractional power:

$${\left(\frac{p^2}{9}\right)}^{-3/2}={\left(\frac{9}{p^2}\right)}^{3/2}=\frac{{\left(3^2\right)}^{3/2}}{{\left(p^2\right)}^{3/2}}=\frac{3^3}{p^3}=3^3{p}^{-3}$$

3. Execute the division operation:

$$\text{LHS}=\frac{2^2\times 3^2}{3^3{p}^{-3}}=2^2\times 3^{2-3}\times p^3=2^2\times 3^{-1}\times p^3$$

Comparing the exponents of the prime bases with the right-hand side ($2^x{3}^y{p}^z$), we find:

* $x=2$
* $y=-1$
* $z=3$