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If the system of equations
$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $
$ x+(\cos \alpha) y+(\sin \alpha) z=0 $
$ x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
$\frac{3 \pi}{4}$
$\frac{7 \pi}{24}$
$\frac{5 \pi}{24}$
$\frac{11 \pi}{24}$
Solution
$\left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \sin \alpha & -\cos \alpha \\ 1 & \cos \alpha & \sin \alpha\end{array}\right|=0$
$ \Rightarrow 1-\sqrt{2} \sin \alpha(\sin \alpha+\cos \alpha)+\sqrt{2} \cos \alpha(\cos \alpha-\sin \alpha)=0 $
$ \Rightarrow 1+\sqrt{2} \cos 2 \alpha-\sqrt{2} \sin 2 \alpha=0 $
$ \cos 2 \alpha-\sin 2 \alpha=-\frac{1}{\sqrt{2}} $
$ \cos \left(2 \alpha+\frac{\pi}{4}\right)=-\frac{1}{2} $
$ 2 \alpha+\frac{\pi}{4}=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} $
$ \alpha+\frac{\pi}{8}=\mathrm{n} \pi \pm \frac{\pi}{3} $
$ \mathrm{n}=0 $
$ \mathrm{x}=\frac{\pi}{3}-\frac{\pi}{8}=\frac{5 \pi}{24}$
