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 \(\frac{2 \pi}{3}\), then \(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=\)

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