Let \(\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}\) 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 \(60^{\circ}\). Then \(|(\bar{a} \times \bar{b}) \times \bar{c}|=\)

\(\overline{ a } \times \overline{ b }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{ i }-2 \hat{ j }+\hat{ k }\)
\(\therefore|\overline{ a } \times \overline{ b }|=\sqrt{4+4+1}=3\). Also \(|\overline{ a }|=\sqrt{4+1+4}=3\)
\(\overline{ c }-\overline{ a }=2 \sqrt{2} \Rightarrow(\overline{ c }-\overline{ a })^2=8\)
\(\therefore \bar{c}-\bar{a}=2 \sqrt{2} \Rightarrow(\bar{c}-\bar{a})^2=8\)
\(\therefore |\overline{ c }|^2+|\overline{ a }|^2=2 \overline{ c } \cdot \overline{ a }=8\)
\(\therefore |\overline{ c }|^2+9-2|\overline{ c }|=8\)
\(\therefore |\overline{ c }|^2-2|\overline{ c }|+1=0 \Rightarrow(|\overline{ c }|-1)^2=0 \Rightarrow|\overline{ c }|=1\)
\(|(\overline{ a } \times \overline{ b }) \times \overline{ c }|=|\overline{ a } \times \overline{ b }| \cdot|\overline{ c }| \cdot \sin 60^{\circ}=(3)(1)\left(\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3}}{2}\)