Two walls of a tunnel lie in the planes 2x + y - z = 3 and x + 3y + 2z = 9, with coordinates in…

Step 1. The point on $z = 0$. The equations become $2x + y = 3$ and $x + 3y = 9$. From the first, $y = 3 - 2x$. Then $x + 9 - 6x = 9$ gives $x = 0$ and $y = 3$. The point is $(0, 3, 0)$.

Step 2. The direction of the edge. $(2, 1, -1) \times (1, 3, 2) = (5, -5, 5)$, which is parallel to $(1, -1, 1)$. The edge is $\dfrac{x}{1} = \dfrac{y - 3}{-1} = \dfrac{z}{1}$.

Step 3. The angle. The dot product of the normals is $2 + 3 - 2 = 3$. The lengths are $\sqrt{6}$ and $\sqrt{14}$. So $\cos\theta = \dfrac{3}{\sqrt{84}} \approx 0.327$ and $\theta \approx 70.89^\circ$.