Let vec{b}=3 hat{i}-2 hat{j}+hat{k} and vec{c | vedclass

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(C) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,we have $\vec{a}=-(\vec{b}+\vec{c})$.Substituting this into the expression $\vec{a} 	imes \vec{b}+\vec{b}

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$\sqrt{234}$

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(C) Given $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,we have $\vec{a}=-(\vec{b}+\vec{c})$.Substituting this into the expression $\vec{a} 	imes \vec{b}+\vec{b} 	imes \vec{c}+\vec{c} 	imes \vec{a}$:$= -(\vec{b}+\vec{c}) 	imes \vec{b}+\vec{b} 	imes \vec{c}+\vec{c} 	imes (-(\vec{b}+\vec{c}))$$= -(\vec{b} 	imes \vec{b}+\vec{c} 	imes \vec{b})+\vec{b} 	imes \vec{c}-(\vec{c} 	imes \vec{b}+\vec{c} 	imes \vec{c})$Since $\vec{b} 	imes \vec{b} = \vec{0}$ and $\vec{c} 	imes \vec{c} = \vec{0}$,and $\vec{c} 	imes \vec{b} = -(\vec{b} 	imes \vec{c})$:$= -(\vec{0} - \vec{b} 	imes \vec{c}) + \vec{b} 	imes \vec{c} - (-(\vec{b} 	imes \vec{c}) + \vec{0})$$= \vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{c} = 3(\vec{b} 	imes \vec{c})$.Now,calculate $\vec{b} 	imes \vec{c}$:$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 1 \ 1 & -1 & -1 \end{vmatrix} = \hat{i}(2+1) - \hat{j}(-3-1) + \hat{k}(-3+2) = 3\hat{i}+4\hat{j}-\hat{k}$.Thus,$3(\vec{b} 	imes \vec{c}) = 3(3\hat{i}+4\hat{j}-\hat{k}) = 9\hat{i}+12\hat{j}-3\hat{k}$.The magnitude is $\sqrt{9^2+12^2+(-3)^2} = \sqrt{81+144+9} = \sqrt{234}$.

Let $\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be two vectors. If $\vec{a}$ is a vector such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then $|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=$

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