Let vec{a} = hat{i} + 2hat{j} - 2hat{k} and v | vedclass

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(E) Step $1$: Calculate $|\vec{a}|$ and $\vec{a} 	imes \vec{b}$.$|\vec{a}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$.$\vec{a} 	imes \vec{b} = \beg

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$3\sqrt{6}/2$

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(E) Step $1$: Calculate $|\vec{a}|$ and $\vec{a} 	imes \vec{b}$.$|\vec{a}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -2 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(2-2) - \hat{j}(1+2) + \hat{k}(-1-2) = 0\hat{i} - 3\hat{j} - 3\hat{k}$.$|\vec{a} 	imes \vec{b}| = \sqrt{0^2 + (-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.Step $2$: Find $|\vec{c}|$.Given $|\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8$.$|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c}) = 8$.Since $\vec{a} \cdot \vec{c} = |\vec{c}|$, let $|\vec{c}| = x$.$x^2 + 9 - 2x = 8 \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies |\vec{c}| = 1$.Step $3$: Calculate $|(\vec{a} 	imes \vec{b}) 	imes \vec{c}|$.$|(\vec{a} 	imes \vec{b}) 	imes \vec{c}| = |\vec{a} 	imes \vec{b}| |\vec{c}| \sin(60^\circ)$.$= (3\sqrt{2}) 	imes (1) 	imes (\sqrt{3}/2) = 3\sqrt{6}/2$.

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