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 \(60^{\circ}\), then the value of \(|(\bar{a} \times \bar{b}) \times \bar{c}|\) is

\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\begin{aligned}
& |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|:=\sqrt{4+4+1}=3 ...(i)\\
& |\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2} \\
& \Rightarrow\left|(\overline{\mathrm{c}}-\overline{\mathrm{a}})^2\right|=8
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow|\vec{c}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=8 \\
& \Rightarrow|\bar{c}|^2+9-2|\bar{c}|=8 \\
& \Rightarrow|\vec{c}|^2-2|\vec{c}|+1=0 \\
& \Rightarrow(|\vec{c}|-1)^2=0 \\
& \Rightarrow|\vec{c}|=1
...(ii)\end{aligned}\)
\(\ldots[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|]\)
Now, \(|(\bar{a} \times \bar{b}) \times \bar{c}|\)
\(=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \cdot|\overline{\mathrm{c}}| \sin 60^{\circ}\)
\(\begin{aligned}
& =(3)(1)\left(\frac{\sqrt{3}}{2}\right) \\
& =\frac{3 \sqrt{3}}{2}
\end{aligned}\)
...[From (i) and (ii)]