The vectors are \(\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}\). If \(\bar{c}\) is a vector such that \(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|\) and \(|\overline{\mathrm{c}}-\overline{\mathrm{a}}|=2 \sqrt{2}\), angle between \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\frac{\pi}{4}\), then \(|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|\) is

Given that angle between \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\frac{\pi}{4}\)
\(\therefore \quad|(\bar{a} \times \bar{b}) \times \bar{c}|=|(\bar{a} \times \bar{b})||\bar{c}| \sin \frac{\pi}{4}\)... (i)
Now, \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|\)
\(\begin{aligned}
& =\hat{\mathrm{i}}(0+2)-\hat{\mathrm{j}}(0+2)+\hat{\mathrm{k}}(2-1) \\
& =2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\
\therefore \quad \mid & |\overline{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{2^2+(-2)^2+1}=3
\end{aligned}\)
Given, \(\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
\(|\vec{a}|=\sqrt{2^2+1^2+(-2)^2}=3\)
Given, \(|\overline{\mathrm{c}}-\mathrm{a}|=2 \sqrt{2}\)
Squaring on both sides, we get
\(\begin{aligned}
& |\overline{\mathrm{c}}|^2+\mid-\mathrm{a}^2-2-\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=8 \\
& \Rightarrow\left|\mathrm{c}^2+3^2-2\right| \mathrm{c} \mid=8 \quad \ldots[\because \cdot \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=|\overline{\mathrm{c}}|]
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow|\bar{c}|^2-2|\vec{c}|+1=0 \\
& \Rightarrow(|\bar{c}|-1)^2=0 \\
& \Rightarrow|\vec{c}|=1
\end{aligned}\)
From (i),
\(\begin{aligned}
|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| & =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})||\overline{\mathrm{c}}| \cdot \sin \frac{\pi}{4} \\
& =3 \times 1 \times \frac{1}{\sqrt{2}} \\
& =\frac{3}{\sqrt{2}}
\end{aligned}\)