Given:
\[ y(x + z) = 19 \quad \text{and} \quad z(x + y) = 51 \]
From the first equation, \( y(x + z) = 19 \). Assuming \( y = 1 \) for simplicity, we get \( x + z = 19 \).
Substitute \( y = 1 \) into the second equation: \( z(x + 1) = 51 \).
Since \( x = 19 - z \), substitute this into the second equation: \( z(19 - z + 1) = 51 \), which simplifies to \( z(20 - z) = 51 \).
Expanding this gives \( 20z - z^2 = 51 \), or \( z^2 - 20z + 51 = 0 \).
Solving the quadratic equation for \( z \):
\[ z = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 51}}{2} = \frac{20 \pm \sqrt{400 - 204}}{2} = \frac{20 \pm \sqrt{196}}{2} = \frac{20 \pm 14}{2} \]
The possible values for \( z \) are:
Now, we calculate \( xyz \) for each case:
If \( z = 3 \), then \( x = 19 - 3 = 16 \), and \( y = 1 \). The product \( xyz = 1 \cdot 16 \cdot 3 = 48 \).
If \( z = 17 \), then \( x = 19 - 17 = 2 \), and \( y = 1 \). The product \( xyz = 1 \cdot 2 \cdot 17 = 34 \).
The minimum value of \( xyz \) is: \[ \boxed{34} \]