Let $\vartriangle ABC$ be a triangle. Let $Q$ be a point in the interior of $\vartriangle ABC$, and let $X, Y,Z$ denote the feet of the altitudes from $Q$ to sides $BC$, $CA$, $AB$, respectively. Suppose that $BC = 15$, $\angle ABC = 60^o$, $BZ = 8$, $ZQ = 6$, and  $\angle QCA = 30^o$. Let line $QX$ intersect the circumcircle of  $\vartriangle XY Z$ at the point $W\ne  X$. If the ratio $\frac{ WY}{WZ}$ can be expressed as $\frac{p}{q}$ for relatively prime positive integers $p, q$, find $p + q$.
1. **Identify the given information and setup the problem:**
   - We have a triangle $\triangle ABC$ with $BC = 15$, $\angle ABC = 60^\circ$, $BZ = 8$, $ZQ = 6$, and $\angle QCA = 30^\circ$.
   - $Q$ is a point inside $\triangle ABC$ and $X, Y, Z$ are the feet of the perpendiculars from $Q$ to $BC$, $CA$, and $AB$ respectively.
   - Line $QX$ intersects the circumcircle of $\triangle XYZ$ at point $W \neq X$.
   - We need to find the ratio $\frac{WY}{WZ}$ and express it as $\frac{p}{q}$ for relatively prime positive integers $p$ and $q$, then find $p + q$.

2. **Use the Law of Sines in $\triangle WXY$ and $\triangle WXZ$:**
   - By the Law of Sines in $\triangle WXY$ and $\triangle WXZ$, we have:
     \[
     \frac{WY}{\sin \angle WXY} = \frac{WX}{\sin \angle WYX}
     \]
     \[
     \frac{WZ}{\sin \angle WXZ} = \frac{WX}{\sin \angle WZX}
     \]
   - Therefore, 
     \[
     \frac{WY}{WZ} = \frac{\sin \angle WXZ}{\sin \angle WXY}
     \]

3. **Relate the angles $\angle WXZ$ and $\angle WXY$ to the given angles:**
   - Since $W$ lies on the circumcircle of $\triangle XYZ$, the angles $\angle WXZ$ and $\angle WXY$ are related to the angles $\angle ZBQ$ and $\angle QCY$ respectively.
   - Specifically, $\angle WXZ = \angle ZBQ$ and $\angle WXY = \angle QCY$.

4. **Calculate the angles $\angle ZBQ$ and $\angle QCY$:**
   - Given $\angle ABC = 60^\circ$ and $BZ = 8$, we can find $\angle ZBQ$ using the fact that $ZQ$ is perpendicular to $AB$.
   - Since $\angle QCA = 30^\circ$, we can directly use this angle.

5. **Use the given lengths and angles to find the ratio:**
   - We know $\angle QCA = 30^\circ$.
   - To find $\angle ZBQ$, we use the fact that $\angle ABC = 60^\circ$ and $BZ = 8$.
   - Since $ZQ$ is perpendicular to $AB$, $\angle ZBQ = 90^\circ - \angle ABC = 90^\circ - 60^\circ = 30^\circ$.

6. **Calculate the ratio using the sine values:**
   - Therefore, $\frac{WY}{WZ} = \frac{\sin 30^\circ}{\sin 30^\circ} = \frac{1/2}{1/2} = 1$.

7. **Express the ratio in the form $\frac{p}{q}$ and find $p + q$:**
   - The ratio $\frac{WY}{WZ} = 1$ can be expressed as $\frac{1}{1}$.
   - Thus, $p = 1$ and $q = 1$.

The final answer is $1 + 1 = \boxed{2}$