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

\(\begin{aligned}
& \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 1 & -2 \\
1 & 1 & 0
\end{array}\right| \\
& =2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\
\therefore \quad & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=\sqrt{4+4+1}=3 ...(i)\\
& |\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2} \\
& \Rightarrow(\overline{\mathrm{c}}-\overline{\mathrm{a}})^2=8
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow|\overline{\mathrm{c}}|^2+|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=8 \\
& \Rightarrow|\overline{\mathrm{c}}|^2+9-2|\overline{\mathrm{c}}|=8 \\
& \Rightarrow|\overline{\mathrm{c}}|^2-2|\overline{\mathrm{c}}|^2+1=0 \\
& \Rightarrow\left(|\overline{\mathrm{c}}|-1\right)^2=0 \\
& \Rightarrow \mid \overline{\mathrm{c}}=1 ...(i)\\
& \text {Now, }|(\overline{\mathrm{a}} \times \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \times \overline{\mathrm{c}}=|\overline{\mathrm{c}}|] \\
& =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \cdot|\overline{\mathrm{c}}| \sin 30^{\circ} \\
& =(3)(1)\left(\frac{1}{2}\right) \\
& =\frac{3}{2}
\end{aligned}\)
...[From (i) and (ii)]