If \(\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{5}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\), then the value of \((2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}\) is

\(\therefore \quad \overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) are perpendicular unit vectors.
\(\begin{aligned}
& \text { Now, }(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\} \\
& =[2 \bar{a}-\bar{b} \bar{a} \times \bar{b} \bar{a}+2 \bar{b}] \\
& =-[\bar{a} \times \bar{b} 2 \bar{a}-\bar{b} \bar{a}+2 \bar{b}] \\
& =-(\bar{a} \times \bar{b}) \cdot\{(2 \bar{a}-\bar{b}) \times(\bar{a}+2 \bar{b})\}
\end{aligned}\)
\(\begin{aligned} & =-(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot 5(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\ & =-5|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=-5|\overline{\mathrm{a}}|^2|\overline{\mathrm{~b}}|^2 \\ & =-5\end{aligned}\)
\(\begin{aligned} & \cdots \cdot[\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}}] \\ & \cdots \cdot[\because|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|=1]\end{aligned}\)