Given bar{a} = 2bar{i} + bar{j} - 2bar{k} and | vedclass

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(C) First,calculate $\bar{a} 	imes \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \bar{i}(0 - (-2))

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$\frac{9}{4}$

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(C) First,calculate $\bar{a} 	imes \bar{b} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 2 & 1 & -2 \ 1 & 1 & 0 \end{vmatrix} = \bar{i}(0 - (-2)) - \bar{j}(0 - (-2)) + \bar{k}(2 - 1) = 2\bar{i} - 2\bar{j} + \bar{k}$.$|\bar{a} 	imes \bar{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.Given $|\bar{c} - \bar{a}|^2 = (2\sqrt{2})^2 = 8$,so $|\bar{c}|^2 + |\bar{a}|^2 - 2(\bar{a} \cdot \bar{c}) = 8$.Since $|\bar{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$ and $\bar{a} \cdot \bar{c} = |\bar{c}|$,we have $|\bar{c}|^2 + 9 - 2|\bar{c}| = 8$,which simplifies to $|\bar{c}|^2 - 2|\bar{c}| + 1 = 0$,so $(|\bar{c}| - 1)^2 = 0$,implying $|\bar{c}| = 1$.Now,$|(\bar{a} 	imes \bar{b}) 	imes \bar{c}| = |\bar{a} 	imes \bar{b}| |\bar{c}| \sin(30^{\circ}) = 3 	imes 1 	imes \frac{1}{2} = \frac{3}{2}$.Therefore,$|(\bar{a} 	imes \bar{b}) 	imes \bar{c}|^2 = (\frac{3}{2})^2 = \frac{9}{4}$.

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