Two walls of a room lie in the planes 2x - y + 2z + 3 = 0 and 3x - 6y + 2z + 14 = 0, with lengths…

Step 1. The bisector. $7(2x - y + 2z + 3) = 3(3x - 6y + 2z + 14)$ gives $14x - 7y + 14z + 21 = 9x - 18y + 6z + 42$. So $5x + 11y + 8z = 21$ and $k = 21$.

Step 2. The angle with the first wall. The normals are $(5, 11, 8)$ and $(2, -1, 2)$. The dot product is $10 - 11 + 16 = 15$. The lengths are $\sqrt{210}$ and $3$. So $\cos\theta = \dfrac{15}{3\sqrt{210}} \approx 0.345$ and $\theta \approx 69.82^\circ$.

Step 3. Acute or obtuse. The dot product of the wall normals is $6 + 6 + 4 = 16 > 0$. With positive constants, the plane with the plus sign bisects the obtuse angle. The walls make an acute angle of about $40.4^\circ$ and an obtuse angle of about $139.6^\circ$. Half of $139.6^\circ$ is about $69.8^\circ$, which agrees.