If $\overrightarrow{a}=-\frac{1}{\sqrt{10}}(3\widehat{i}+\widehat{k}),\overrightarrow{b}=\frac{1}{7}(2\widehat{i}+3\widehat{j}-6\overrightarrow{k})$, then the value of 

$(2\overrightarrow{a}-\overrightarrow{b})\cdot \left\{\right(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{a}+2\overrightarrow{b}\left)\right\}$, is 

Clearly, $\overrightarrow{a}$and $\overrightarrow{b}$ are perpendicular unit vectors.

 Now,

$\begin{array}{l}(2\overrightarrow{a}-\overrightarrow{b})\cdot \left\{\right(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{a}+2\overrightarrow{b}\left)\right\}\\ =[2\overrightarrow{a}-\overrightarrow{b},\overrightarrow{a}\times \overrightarrow{b},\overrightarrow{a}+2\overrightarrow{b}]\\ =-[\overrightarrow{a}\times \overrightarrow{b},2\overrightarrow{a}-\overrightarrow{b},\overrightarrow{a}+2\overrightarrow{b}]\\ =-(\overrightarrow{a}\times \overrightarrow{b})\cdot \left\{\right(2\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+2\overrightarrow{b}\left)\right\}\end{array}$

$=-(\overrightarrow{a}\times \overrightarrow{b})\cdot 5(\overrightarrow{a}\times \overrightarrow{b})$

$=-5|\overrightarrow{a}\times \overrightarrow{b}|=-5\left|\overrightarrow{a}{|}^2\right|\overrightarrow{b}{|}^2[\because \overrightarrow{a}\perp \overrightarrow{b}]$

$=-5[\because |\overrightarrow{a}|=|\overrightarrow{b}|=1]$