If the system of equations

$$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$$

has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

<p>$$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$$</p> <p>$\because$ Non-trivial solution</p> <p>$\Rightarrow D=0$</p> <p>$$\begin{aligned} & \left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\ \end{aligned}$$</p> <p>$$\begin{aligned} & -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\ & \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\ & \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\ & \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\ & \Rightarrow \alpha=\frac{5 \pi}{24} \end{aligned}$$</p>