Let \(\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}\). If \(\overline{\mathrm{c}}\) 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 \(\frac{\pi}{6}\), then \(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|\) is

\(
\begin{aligned}
& |\overline{\mathrm{c}}-\overline{\mathrm{a}}|^2=2 \sqrt{2} \\
& \Rightarrow|\overline{\mathrm{c}}|^2+|\overline{\mathrm{a}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=8 \\
& \Rightarrow|\overline{\mathrm{c}}|^2+9-2|\overline{\mathrm{c}}|=8 \\
& \Rightarrow\left(|\overline{\mathrm{c}}|^{-1}\right)^2=0 \\
& \Rightarrow|\overline{\mathrm{c}}|^2=1
\end{aligned}
\)
Now,
\(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})||\overline{\mathrm{c}}| \sin \frac{\pi}{6} \)
\( =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|(1)\left(\frac{1}{2}\right) \quad \ldots[\text { From (i) }] \)
\( =\frac{3}{2} \quad \ldots[\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}]\)