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

Since \(\bar{a} \cdot \bar{b}=0\)
\(\therefore \quad \overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) are perpendicular unit vectors.
Now, \((2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})]\)
\(\begin{aligned} & =[2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}} \times \overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-[\overline{\mathrm{a}} \times \overline{\mathrm{b}} 2 \overline{\mathrm{a}}-\overline{\mathrm{b}} \overline{\mathrm{a}}+2 \overline{\mathrm{b}}] \\ & =-(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{(2 \overline{\mathrm{a}}-\overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+2 \overline{\mathrm{b}})\} \\ & =-5(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})\end{aligned}\)
\(\begin{array}{ll}=-5|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=-5|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2 & \ldots . \cdot[\because \overline{\mathrm{a}} \perp \overline{\mathrm{b}}] \\ =-5 & \ldots . \cdot[\because|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|=1]\end{array}\)