$ 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$
ને એક અસામાન્ય ઉકેલ હોય, તો $\alpha \in\left(0, \frac{\pi}{2}\right)$ બરાબર ............ છે.
$ \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}$