Let \(\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}\) and \(\bar{c}\) be a vector such that \(|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=4, \quad|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=3\) and the angle between \(\overline{\mathrm{c}}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}\) is \(\frac{\pi}{6}\), then \(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\) is equal to

\(\begin{array}{ll} & \overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ \therefore \quad & |\overline{\mathrm{a}}|=\sqrt{4+1+4}=3 \\ & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=\sqrt{4+4+1}=3\end{array}\)
\(\text { Angle between } \overline{\mathrm{c}} \text { and } \overline{\mathrm{a}} \times \overline{\mathrm{b}} \text { is } \frac{\pi}{6} \quad \ldots \text { [Given] }\)
\(\begin{aligned} \therefore \quad & \sin \frac{\pi}{6}=\frac{|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}} \| \overline{\mathrm{c}}|} \\ & \frac{1}{2}=\frac{3}{3 \times|\overline{\mathrm{c}}|} \Rightarrow|\overline{\mathrm{c}}|=2\end{aligned}\)
Now, \(|\bar{c}-\bar{a}|=4\).. [Given]
\(\begin{aligned} & \Rightarrow|\bar{c}|^2+|\bar{a}|^2-2 \bar{a} \cdot \bar{c}=16 \\ & \Rightarrow 4+9-2 a \cdot c=16 \\ & \Rightarrow a \cdot c=\frac{-3}{2}\end{aligned}\)