If \(\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{\imath}+\hat{k}), \overline{\mathrm{b}}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})\), then the value of
\((2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})]\) is

Given vectors \(\bar{a} \& \bar{b}\) are unit vectors.
\(\text {Now }(2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})] \)
\( =-(2 \bar{a}-\bar{b}) \cdot[(\bar{a}+2 \bar{b}) \times(\bar{a} \times \bar{b})] \)
\( =-(2 \bar{a}-\bar{b}) \cdot[\bar{a} \times(\bar{a} \times \bar{b})+2 \bar{b} \times(\bar{a} \times \bar{b})] \)
\( =-(2 \bar{a}-\bar{b})[(\bar{a} \cdot \bar{b}) \cdot \bar{a}-(\bar{a} \cdot \bar{a}) \cdot \bar{b}+2(\bar{b} \cdot \bar{b}) \cdot \bar{a}-2\)\((\bar{b} \cdot \bar{a}) \cdot \bar{b}] \)
\( =-(2 \bar{a}-\bar{b})[(\bar{a} \cdot \bar{b}) \bar{a}-\bar{b}+2 \bar{a}-2(\bar{b} \cdot \bar{a}) \bar{b}] \)
\( \text {Here } \bar{a} \cdot \bar{b}=\frac{3(2)}{7 \sqrt{10}}-\frac{1(6)}{7 \sqrt{10}}=0\)
Hence given expression becomes
\(\begin{array}{l}
=-(2 \bar{a}-\bar{b})[-\bar{b}+2 \bar{a}]=-(2 \bar{a}-\bar{b})^{2} \\
=-\left[4 \bar{a}^{2}+\bar{b}^{2}-4 \bar{a} \cdot \bar{b}\right]=-(4+1)=-5
\end{array}\)