If $\overline{a}$ and $\overline{\mathrm{b}}$ are vectors in space given by $\stackrel{\mathrm{‾}}{\mathrm{a}}=\frac{\stackrel{\mathrm{‾}}{\mathrm{i}}-2\stackrel{\mathrm{‾}}{\mathrm{j}}}{\sqrt{5}},\stackrel{\mathrm{‾}}{\mathrm{b}}=\frac{2\stackrel{\mathrm{‾}}{\mathrm{i}}+\stackrel{\mathrm{‾}}{\mathrm{j}}+3\stackrel{\mathrm{‾}}{\mathrm{k}}}{\sqrt{14}}$  then value of
$(2\stackrel{\mathrm{‾}}{\mathrm{a}}+\stackrel{\mathrm{‾}}{\mathrm{b}})\cdot \left[\right(\stackrel{\mathrm{‾}}{\mathrm{a}}\times \stackrel{\mathrm{‾}}{\mathrm{b}})\times (\stackrel{\mathrm{‾}}{\mathrm{a}}-2\stackrel{\mathrm{‾}}{\mathrm{b}}\left)\right]$ is

$$\begin{gathered} \left|\overline{a}\right|=\left|\overline{b}\right|=1,\overline{a}.\overline{b}=0 \\ (\stackrel{\mathrm{‾}}{\mathrm{a}}\times \stackrel{\mathrm{‾}}{\mathrm{b}})\times (\stackrel{\mathrm{‾}}{\mathrm{a}}-2\stackrel{\mathrm{‾}}{\mathrm{b}})=\left(\overline{a}.\left(\overline{a}-2\overline{b}\right)\right)\overline{b}-\left(\overline{b}.\left(\overline{a}-2\overline{b}\right)\right).\overline{a}=\overline{b}+2\overline{a} \\ \therefore (2\stackrel{\mathrm{‾}}{\mathrm{a}}+\stackrel{\mathrm{‾}}{\mathrm{b}})\cdot \left[\right(\stackrel{\mathrm{‾}}{\mathrm{a}}\times \stackrel{\mathrm{‾}}{\mathrm{b}})\times (\stackrel{\mathrm{‾}}{\mathrm{a}}-2\stackrel{\mathrm{‾}}{\mathrm{b}}\left)\right]=\left(2\overline{a}+\overline{b}\right).\left(\overline{b}+2\overline{a}\right) \\ ={\left|2\overline{a}+\overline{b}\right|}^2=4{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+4\overline{a}.\overline{b}=4{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2=5 \end{gathered}$$