**Solution**

Given: p, q, and r are in AP and x, y, and z are in GP.

We know that if a, b, and c are in GP, then $b^2 = ac$, and if a, b, and c are in AP, then $b = \frac{a+c}{2}$.

Let's use these formulas to simplify the expression we need to find.

**Simplifying the Expression**

We are given the expression $x^{q-r} y^{r-p} z^{p-q}$.

We know that p, q, and r are in AP, so we can write:

$q = \frac{p+r}{2}$

$r = \frac{q+p}{2}$

We can substitute these expressions for q and r in the expression we need to find:

$x^{q-r} y^{r-p} z^{p-q} = x^{\frac{p+r}{2} - \frac{q+p}{2}} y^{\frac{q+p}{2} - p} z^{p-\frac{p+r}{2}}$

Simplifying the exponents, we get:

$x^{\frac{r-p}{2}} y^{\frac{q-p}{2}} z^{\frac{p-r}{2}}$

Now, we can use the fact that x, y, and z are in GP to write:

$y^2 = xz$

We can substitute this expression for y^2 in the expression we need to find:

$x^{\frac{r-p}{2}} y^{\frac{q-p}{2}} z^{\frac{p-r}{2}} = x^{\frac{r-p}{2}} (xz)^{\frac{q-p}{4}} z^{\frac{p-r}{2}}$

Simplifying, we get:

$x^{\frac{r-p}{2}} x^{\frac{q-p}{4}} z^{\frac{q-p}{4}} z^{\frac{p-r}{2}}$

Combining the exponents, we get:

$x^{\frac{3q-5p+r}{4}} z^{\frac{3p-5r+q}{4}}$

**Final Answer**

Therefore, $x^{q-r} y^{r-p} z^{p-q} = x^{\frac{3q-5p+r}{4}} z^{\frac{3p-5r+q}{4}}$.