The sum of all possible values of $$\theta \in [0, 2\pi]$$, for which the system of equations :
$$x\cos 3\theta - 8y - 12z = 0$$
$$x\cos 2\theta + 3y + 3z = 0$$
$$x + y + 3z = 0$$
has a non-trivial solution, is equal to :

$$\Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$$

$$\Delta = \begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0$$

$$R_2 \to R_2 - R_3$$:

$$\Delta = \begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta - 1 & 2 & 0 \\ 1 & 1 & 3 \end{vmatrix} = 0$$

$$R_1 \to R_1 + 4R_3$$

$$\Delta = \begin{vmatrix} \cos 3\theta + 4 & -4 & 0 \\ \cos 2\theta - 1 & 2 & 0 \\ 1 & 1 & 3 \end{vmatrix} = 0$$

$$\Delta = 3 \cdot \begin{vmatrix} \cos 3\theta + 4 & -4 \\ \cos 2\theta - 1 & 2 \end{vmatrix} = 0$$

$$2(\cos 3\theta + 4) - (-4)(\cos 2\theta - 1) = 0$$

$$\cos 3\theta + 2\cos 2\theta + 2 = 0$$

$$(4\cos^3 \theta - 3\cos \theta) + 2(2\cos^2 \theta - 1) + 2 = 0$$

$$\cos \theta \cdot (4\cos^2 \theta + 4\cos \theta - 3) = 0$$

$$\cos \theta \cdot (2\cos \theta - 1) \cdot (2\cos \theta + 3) = 0$$

$$\theta = \frac{\pi}{2}, \ \frac{3\pi}{2}$$

$$\theta = \frac{\pi}{3}, \ \left(2\pi - \frac{\pi}{3}\right) = \frac{5\pi}{3}$$

$$\text{Sum} = \frac{\pi}{2} + \frac{3\pi}{2} + \frac{\pi}{3} + \frac{5\pi}{3}$$

$$\text{Sum} = \left(\frac{4\pi}{2}\right) + \left(\frac{6\pi}{3}\right) = 2\pi + 2\pi = 4\pi$$