A tetrahedron is a polyhedron composed of four triangular faces. Faces $ABC$ and $BCD$ of a tetrahedron $ABCD$ meet at an angle of $\pi/6$. The area of triangle $\triangle ABC$ is $120$. The area of triangle $\triangle BCD$ is $80$, and $BC = 10$. What is the volume of the tetrahedron? We call the volume of a tetrahedron as one-third the area of it's base times it's height.
1. **Determine the height \( h_D \) of triangle \( \triangle BCD \):**
   Given the area of \( \triangle BCD \) is 80 and the base \( BC = 10 \), we can use the formula for the area of a triangle:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Substituting the given values:
   \[
   80 = \frac{1}{2} \times 10 \times h_D
   \]
   Solving for \( h_D \):
   \[
   80 = 5 \times h_D \implies h_D = \frac{80}{5} = 16
   \]

2. **Calculate the height \( H \) of the tetrahedron:**
   The height \( H \) of the tetrahedron is the perpendicular distance from vertex \( A \) to the plane containing \( \triangle BCD \). Given that the angle between the planes of \( \triangle ABC \) and \( \triangle BCD \) is \( \frac{\pi}{6} \), we can use trigonometry to find \( H \):
   \[
   H = h_D \cdot \sin \left( \frac{\pi}{6} \right)
   \]
   Since \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \):
   \[
   H = 16 \cdot \frac{1}{2} = 8
   \]

3. **Calculate the volume of the tetrahedron:**
   The volume \( V \) of a tetrahedron is given by:
   \[
   V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
   \]
   Here, the base area is the area of \( \triangle ABC \), which is 120, and the height \( H \) is 8:
   \[
   V = \frac{1}{3} \times 120 \times 8
   \]
   Simplifying:
   \[
   V = \frac{1}{3} \times 960 = 320
   \]

The final answer is \( \boxed{320} \).