Let a vector $\overline{V_1}=\overline{OP},$ Where O is origin and P is (2,0,2). Vector $\overline{V_1}$ is rotated about origin by an angle of $\frac{\pi }{3}$  in a plane which is perpendicular to XOZ plane to get vector $\overline{V_2}$ , where $\overline{V_2}.\stackrel{\wedge }{j}>0$. Again  $\overline{V_1}$ is rotated about origin by right angle in XOZ plane to get vector $\overline{V_3}$. If V is volume of tetrahedron whose coterminous edges are $\overline{V_1},\overline{V_2}$   and $\overline{V_3}$  then value of $\frac{3V}{\sqrt{6}}$  is

Equation of plane perpendicular to XOZ plane is $P:x-z=0$ 
$\overrightarrow{V_2}$ is coplanar  $\overrightarrow{V_1}$ and j

 
 
 
 $\begin{array}{l}\Rightarrow {\overrightarrow{V}}_2=x{\overrightarrow{V}}_1+y\stackrel{\wedge }{j}\\ \Rightarrow {\overrightarrow{V}}_2=x(2\stackrel{\wedge }{i}+2\stackrel{\wedge }{k})+y\stackrel{\wedge }{j}\\ \because \left|{\overrightarrow{V}}_2\right|=\left|{\overrightarrow{V}}_1\right|=2\sqrt{2}\\ \therefore 4x^2+y^2+4x^2=8\end{array}$
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Now,  ${\overrightarrow{V}}_2.{\overrightarrow{V}}_1=\left|{\overrightarrow{V}}_2\right|\left|{\overrightarrow{V}}_1\right|\cos \frac{\pi }{3}=4$$4=(x(2\stackrel{\wedge }{i}+2\stackrel{\wedge }{k})+y\stackrel{\wedge }{j}).(2\stackrel{\wedge }{i}+2\stackrel{\wedge }{k})$

$\Rightarrow {\overrightarrow{V}}_2=\stackrel{\wedge }{i}+\sqrt{6}\stackrel{\wedge }{j}+\stackrel{\wedge }{k}(\because {\overrightarrow{V}}_2.\stackrel{\wedge }{j}>0)$

 
Now, ${\overrightarrow{V}}_3={\overrightarrow{V}}_1+x\stackrel{\wedge }{i}$

$\because {\overrightarrow{V}}_3.{\overrightarrow{V}}_1=0\Rightarrow 8+2x=0\Rightarrow x=-4$
 $\Rightarrow {\overrightarrow{V}}_3=2\stackrel{\wedge }{i}+2\stackrel{\wedge }{k}-4\stackrel{\wedge }{i}$  ${\overrightarrow{V}}_3=-2\stackrel{\wedge }{i}+2\stackrel{\wedge }{k}$
 
$\Rightarrow$ Volume of tetrahedron 
 

$\Rightarrow V=\frac{1}{6}8\sqrt{6}\Rightarrow \frac{3V}{\sqrt{6}}=4$
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