The vectors are bar{a}=2 hat{i}+hat{j}-2 hat{ | vedclass

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(B) Given that the angle between $\bar{a} 	imes \bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$.Using the definition of the cross product magnitude,$|(\bar

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$\frac{3}{\sqrt{2}}$

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(B) Given that the angle between $\bar{a} 	imes \bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$.Using the definition of the cross product magnitude,$|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = |\bar{a} 	imes \bar{b}| |\bar{c}| \sin(\frac{\pi}{4})$. ... $(i)$First,calculate $\bar{a} 	imes \bar{b}$:$\bar{a} 	imes \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.The magnitude is $|\bar{a} 	imes \bar{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Given $|\bar{c} - \bar{a}| = 2\sqrt{2}$. Squaring both sides:$|\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = (2\sqrt{2})^2 = 8$.Since $|\bar{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$ and $\bar{a} \cdot \bar{c} = |\bar{c}|$,we have:$|\bar{c}|^2 + 3^2 - 2|\bar{c}| = 8$.$|\bar{c}|^2 - 2|\bar{c}| + 9 = 8 \implies |\bar{c}|^2 - 2|\bar{c}| + 1 = 0$.$(|\bar{c}| - 1)^2 = 0 \implies |\bar{c}| = 1$.Substituting these values into equation $(i)$:$|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = 3 	imes 1 	imes \sin(\frac{\pi}{4}) = 3 	imes \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.

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