Answer:
Option C
Explanation:
$x+\sin \alpha y+\cos \alpha z=0, x+ \cos \alpha y+ \sin \alpha z=0$
$-x+\sin \alpha y-\cos \alpha z=0$
Non-trivial sol, so
$\begin{bmatrix}1 & \sin \alpha& \cos \alpha \\1 & \cos \alpha& \sin \alpha\\ -1 & \sin \alpha & -\cos \alpha \end{bmatrix}=0$
$\Rightarrow$ $| 1(-\sin^{2} \alpha -\cos^{2} \alpha)-\sin \alpha(-\cos \alpha+\sin \alpha)+ \cos \alpha( \sin \alpha+\cos \alpha)|$=0
$ -1- \sin \alpha (- \cos \alpha+ \sin \alpha] +\cos \alpha [ \sin \alpha+\cos \alpha]=0$
$-1+ \sin \alpha \cos \alpha- \sin^{2} \alpha+ \sin \alpha \cos \alpha + \cos ^{2} \alpha=0$
$-1+ \sin 2 \alpha+\cos 2 \alpha=0$
$\sin 2 \alpha+\cos 2 \alpha=1$
$\Rightarrow$ $\frac{1}{\sqrt{2}}\sin\alpha+\frac{1}{\sqrt{2}} \cos\alpha=\frac{1}{\sqrt{2}}$
$\Rightarrow$ $\sin\left(2\alpha+\frac{\pi}{4}\right)=\sin\frac{\pi}{4}$
$\Rightarrow$ $2\alpha+\frac{\pi}{4}=n\pi+(-1)^{n}\frac{\pi}{4}$
$2\alpha=n\pi+(-1)^{n}\frac{\pi}{4}-\frac{\pi}{4}$
$\Rightarrow$ $\alpha$= $\frac{n \pi}{2}+(-1)^{n} \frac{\pi}{8}-\frac{\pi}{8}$
