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