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

$\overrightarrow{\mathrm{b}}$ perpendicular $\overrightarrow{\mathrm{c}}\Rightarrow \overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=0\Rightarrow {\tan }^2\mathrm{\alpha }-\tan \mathrm{\alpha }-6=0\Rightarrow \tan \mathrm{\alpha }=3,-2$

Also $\overrightarrow{\mathrm{a}}$make an obtuse angle with z-axis.

Therefore, $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{k}}<0\Rightarrow \sin 2\mathrm{\alpha }<0$

If $\tan \mathrm{\alpha }=3,$ then $\sin 2\mathrm{\alpha }=\frac{2\tan \mathrm{\alpha }}{1+{\tan }^2\mathrm{\alpha }}=\frac{6}{10}>0$

Threfore, $\tan \mathrm{\alpha }=-2\Rightarrow \tan (\mathrm{\pi }-\mathrm{\alpha })=2$

$\Rightarrow \mathrm{\alpha }=(2\mathrm{n}+1)\mathrm{\pi }-{\tan }^{-1}2,\mathrm{n}\in \mathrm{I}$