Let \(\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}=\hat{i}+\hat{\mathbf{j}}\). Let \(\overline{\mathrm{c}}\) be a vector such that \(|\bar{c}-\bar{a}|=3\) and \(|(\bar{a} \times \bar{b}) \times \bar{c}|=3\) and the angle between \(\overline{\mathrm{c}}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}\) is \(30^{\circ}\), then \(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\) is equal to

\(\begin{array}{ll} & \bar{a}=2 \hat{i}+\hat{j}-2 \hat{k} \\ & \bar{b}=\hat{i}+\hat{j} \\ & |\bar{a}|=\sqrt{4+1+4}=3 \\ & \bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\hat{k} \\ \therefore \quad & |\bar{a} \times \bar{b}|=\sqrt{4+4+1}=3\end{array}\)
Angle between \(\overline{\mathrm{c}}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=30^{\circ}\)
...[Given]
\(\sin 30^{\circ}=\frac{|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}||\overline{\mathrm{c}}|}\)
\(\begin{aligned}
& \frac{1}{2}=\frac{3}{3 \times|\overline{\mathrm{c}}|} \\
& \Rightarrow|\overline{\mathrm{c}}|=2
\end{aligned}\)
\(\begin{aligned}
& \text { Now, }|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=3 \\
& \Rightarrow|\overline{\mathrm{c}}|^2+|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=9 \\
& \Rightarrow 4+9-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=9 \\
& \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=2
\end{aligned}\)