If vectors $\overrightarrow{b}=(\tan \alpha ,-1,2\sqrt{\sin \alpha /2})$ and $\overrightarrow{c}=\left(\tan \alpha ,\tan \alpha ,-\frac{3}{\sqrt{\sin \alpha /2}}\right)$ are orthogoanl and vector $\overrightarrow{a}=(1,3,\sin 2\alpha )$ makes an obtuse angle with the z–axis, then the value of $\alpha$ is

Since $\overrightarrow{a}=(1,3,\sin 2\alpha )$ makes an obtuse angle with the z–axis, its z–component is negative.

Thus, $-1\le \sin 2\alpha <0\dots \left(i\right)$

But $\overrightarrow{b}\cdot \overrightarrow{c}=0(\because \text{ orthogonal })$

${\tan }^2\alpha -\tan \alpha -6=0$

$\therefore (\tan \alpha -3)(\tan \alpha +2)=0\Rightarrow \tan \alpha =3,-2$

Now, $\tan \alpha =3$

$\therefore \sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{6}{1+9}=\frac{3}{5}$

(not possible as $\sin 2\alpha <0$)

Now, if $\tan \alpha =-2$ ,

$\Rightarrow \sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{-4}{1+4}=\frac{-4}{5}\Rightarrow \tan 2\alpha >0$

Hence, $2\alpha$ is the third quadrant. Also, $\sqrt{\sin \alpha /2}$ is meaningful. If $0<\sin \alpha /2<1\text{, then }$ $\alpha =(4n+1)\pi -{\tan }^{-1}2\text{ and }\alpha =(4n+2)\pi -{\tan }^{-1}2$