Determine the number of positive integer solutions $(x,y, z)$ to the equation $xyz = 2(x + y + z)$.
1. We start with the given equation:
   \[
   xyz = 2(x + y + z)
   \]

2. Assume \(x, y, z\) are positive integers. We first note that \(x, y, z\) cannot all be the same. If they were, we would have:
   \[
   x^3 = 6x \implies x^2 = 6 \implies x = \sqrt{6}
   \]
   Since \(x\) must be an integer, this is not possible.

3. Without loss of generality, assume \(x \leq y \leq z\). This implies:
   \[
   xyz \leq 6z \implies xy \leq 6
   \]

4. We now check all possible pairs \((x, y)\) such that \(xy \leq 6\):
   - \(x = 1, y = 1\)
   - \(x = 1, y = 2\)
   - \(x = 1, y = 3\)
   - \(x = 1, y = 4\)
   - \(x = 1, y = 5\)
   - \(x = 1, y = 6\)
   - \(x = 2, y = 2\)
   - \(x = 2, y = 3\)
   - \(x = 2, y = 4\)
   - \(x = 3, y = 3\)

5. For each pair \((x, y)\), we solve for \(z\) using the original equation:
   \[
   z = \frac{2(x + y + z)}{xy}
   \]
   Rearranging, we get:
   \[
   z(xy - 2) = 2(x + y) \implies z = \frac{2(x + y)}{xy - 2}
   \]

6. We now check each pair:
   - For \((x, y) = (1, 1)\):
     \[
     z = \frac{2(1 + 1)}{1 \cdot 1 - 2} = \frac{4}{-1} \quad \text{(not a positive integer)}
     \]
   - For \((x, y) = (1, 2)\):
     \[
     z = \frac{2(1 + 2)}{1 \cdot 2 - 2} = \frac{6}{0} \quad \text{(undefined)}
     \]
   - For \((x, y) = (1, 3)\):
     \[
     z = \frac{2(1 + 3)}{1 \cdot 3 - 2} = \frac{8}{1} = 8 \quad \text{(valid solution)}
     \]
   - For \((x, y) = (1, 4)\):
     \[
     z = \frac{2(1 + 4)}{1 \cdot 4 - 2} = \frac{10}{2} = 5 \quad \text{(valid solution)}
     \]
   - For \((x, y) = (1, 5)\):
     \[
     z = \frac{2(1 + 5)}{1 \cdot 5 - 2} = \frac{12}{3} = 4 \quad \text{(valid solution)}
     \]
   - For \((x, y) = (1, 6)\):
     \[
     z = \frac{2(1 + 6)}{1 \cdot 6 - 2} = \frac{14}{4} = 3.5 \quad \text{(not an integer)}
     \]
   - For \((x, y) = (2, 2)\):
     \[
     z = \frac{2(2 + 2)}{2 \cdot 2 - 2} = \frac{8}{2} = 4 \quad \text{(valid solution)}
     \]
   - For \((x, y) = (2, 3)\):
     \[
     z = \frac{2(2 + 3)}{2 \cdot 3 - 2} = \frac{10}{4} = 2.5 \quad \text{(not an integer)}
     \]
   - For \((x, y) = (2, 4)\):
     \[
     z = \frac{2(2 + 4)}{2 \cdot 4 - 2} = \frac{12}{6} = 2 \quad \text{(valid solution)}
     \]
   - For \((x, y) = (3, 3)\):
     \[
     z = \frac{2(3 + 3)}{3 \cdot 3 - 2} = \frac{12}{7} \quad \text{(not an integer)}
     \]

7. The valid solutions are:
   - \((x, y, z) = (1, 3, 8)\)
   - \((x, y, z) = (1, 4, 5)\)
   - \((x, y, z) = (1, 5, 4)\)
   - \((x, y, z) = (2, 2, 4)\)
   - \((x, y, z) = (2, 4, 2)\)

The final answer is \( \boxed{ (1, 3, 8), (1, 4, 5), (1, 5, 4), (2, 2, 4), (2, 4, 2) } \)