If the system of equationsx+(sqrt{2} sin alph | vedclass

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(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1

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$\frac{5 \pi}{24}$

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(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ 1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0$Expanding along the first row:$1(-\cos^2 \alpha - \sin^2 \alpha) - \sqrt{2} \sin \alpha(-\cos \alpha - \sin \alpha) + \sqrt{2} \cos \alpha(\sin \alpha - \cos \alpha) = 0$$-1 + \sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} \sin^2 \alpha + \sqrt{2} \sin \alpha \cos \alpha - \sqrt{2} \cos^2 \alpha = 0$$-1 + 2\sqrt{2} \sin \alpha \cos \alpha - \sqrt{2}(\cos^2 \alpha - \sin^2 \alpha) = 0$$-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0$$\sqrt{2}(\sin 2\alpha - \cos 2\alpha) = 1$$\sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$$\frac{1}{\sqrt{2}} \sin 2\alpha - \frac{1}{\sqrt{2}} \cos 2\alpha = \frac{1}{2}$$\sin(2\alpha - \frac{\pi}{4}) = \frac{1}{2}$Since $\alpha \in (0, \frac{\pi}{2})$,then $2\alpha \in (0, \pi)$,so $2\alpha - \frac{\pi}{4} \in (-\frac{\pi}{4}, \frac{3\pi}{4})$.Thus,$2\alpha - \frac{\pi}{4} = \frac{\pi}{6}$ or $2\alpha - \frac{\pi}{4} = \frac{5\pi}{6}$.Case $1$: $2\alpha = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \Rightarrow \alpha = \frac{5\pi}{24}$.Case $2$: $2\alpha = \frac{5\pi}{6} + \frac{\pi}{4} = \frac{13\pi}{12} \Rightarrow \alpha = \frac{13\pi}{24}$ (Outside the range).Therefore,$\alpha = \frac{5\pi}{24}$.

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 :

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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 :