If \(\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})\) and \(\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})\), then the value of \((2 \hat{a}-\hat{b}) \cdot[(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})]\) is

Here, \(\hat{a} \cdot \hat{b}=0\)
\(\therefore \quad \hat{a}\) and \(\hat{b}\) are perpendicular unit vectors.
Now, \((2 \hat{a}-\hat{b}) \cdot\{(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})\}\)
\(=\left[\begin{array}{lll}
2 \hat{a}-\hat{b} & \hat{a} \times \hat{b} & \hat{a}+2 \hat{b}
\end{array}\right]\)
\(\begin{aligned} & =-[\hat{a} \times \hat{b} \quad 2 \hat{a}-\hat{b} \quad \hat{a}+2 \hat{b}] \\ & =-(\hat{a} \times \hat{b}) \cdot\{(2 \hat{a}-\hat{b}) \times(\hat{a}+2 \hat{b})\} \\ & =-(\hat{a} \times \hat{b}) \cdot 5(\hat{a} \times \hat{b}) \\ & =-5|\hat{a} \times \hat{b}|=-5|\hat{a}|^2|\hat{b}|^2 \\ & =-5\end{aligned}\)
\(\begin{aligned} & \ldots[\because \hat{a} \perp \hat{b}] \\ & \ldots[\because|\hat{a}|=|\hat{b}|=1]\end{aligned}\)