If $216^{2/3}÷{\left(\frac{p^2}{9}\right)}^{-3/2}=2^x{3}^y{p}^z$, then the values of $x$, $y$, and $z$ are:

Let us simplify the left-hand side (LHS) of the given equation step-by-step:
First, express 216 as a power of prime bases:

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

Now apply the exponent $\frac{2}{3}$:

$$216^{2/3}={\left(6^3\right)}^{2/3}=6^{3\times \frac{2}{3}}=6^2=36$$

In terms of prime factors:

$$36=2^2\times 3^2$$

Next, simplify the second term containing $p$:

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

Now substitute these back into the LHS division operation:

$$\text{LHS}=216^{2/3}÷{\left(\frac{p^2}{9}\right)}^{-3/2}$$

$$\text{LHS}=(2^2\times 3^2)÷(3^3\times p^{-3})$$

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

We are given that:

$$\text{LHS}=2^x{3}^y{p}^z$$

$$2^2\times 3^{-1}\times p^3=2^x{3}^y{p}^z$$

Comparing exponents on both sides, we get:

$$x=2$$

$$y=-1$$

$$z=3$$

This matches option A.