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 \(30^{\circ}\), then the value of \(|(\bar{a} \times \bar{b}) \times \bar{c}|\) is equal to

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