Given the angle between planes $\pi_1$ and $\pi_2$ is equal to the angle between...

Question

I'm not sure where I'm going wrong with this question but i keep coming to a hexic equation rather than a quartic equation.

the three planes: $$\pi_1: ax+2y+z=3$$ $$\pi_2: x+ay+z=4$$ $$\pi_3: x+y+az=5$$

Given the angle between planes $\pi_1$ and $\pi_2$ is equal to the angle between $\pi_2$ and $\pi_3$, show that $a$ must statisfy the quartic equation $$5a^4+2a^3-2a^2-8a-3=0$$

Answer 1

Given the planes

$$ \pi_k \to \left<p-p_k,\vec n_k\right> = 0 $$

with $p = (x,y,z)$

if

$$ \frac{\left< \vec n_1,\vec n_2\right> ^2}{||\vec n_1||^2||\vec n_2||^2} = \frac{\left< \vec n_2,\vec n_3\right> ^2}{||\vec n_2||^2||\vec n_3||^2} $$

then

$$ 5a^4+2a^3-2a^2-8a-3 = 0 $$

here $\left<\cdot,\cdot\right>$ represents the scalar product of two vectors