Let vec{a}=6 hat{i}+9 hat{j}+12 hat{k},vec{b} | vedclass

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(B) Given $\vec{a} 	imes \vec{c} = \vec{a} 	imes \vec{b}$,we have $\vec{a} 	imes (\vec{c} - \vec{b}) = \vec{0}$.This implies that $(\vec{c} - \vec{

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$11$

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(B) Given $\vec{a} 	imes \vec{c} = \vec{a} 	imes \vec{b}$,we have $\vec{a} 	imes (\vec{c} - \vec{b}) = \vec{0}$.This implies that $(\vec{c} - \vec{b})$ is parallel to $\vec{a}$.Thus,$\vec{c} - \vec{b} = k \vec{a}$ for some scalar $k$,or $\vec{c} = \vec{b} + k \vec{a}$.Substituting $\vec{c} = (\alpha + 6k) \hat{i} + (11 + 9k) \hat{j} + (-2 + 12k) \hat{k}$ into $\vec{a} \cdot \vec{c} = -12$:$6(\alpha + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12$.$6\alpha + 36k + 99 + 81k - 24 + 144k = -12 \Rightarrow 6\alpha + 261k = -87$.Using $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$:$(\alpha + 6k) - 2(11 + 9k) + (-2 + 12k) = 5$.$\alpha + 6k - 22 - 18k - 2 + 12k = 5 \Rightarrow \alpha = 29$.Substituting $\alpha = 29$ into $6(29) + 261k = -87$:$174 + 261k = -87 \Rightarrow 261k = -261 \Rightarrow k = -1$.Thus,$\vec{c} = \vec{b} - \vec{a} = (29-6)\hat{i} + (11-9)\hat{j} + (-2-12)\hat{k} = 23\hat{i} + 2\hat{j} - 14\hat{k}$.Finally,$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$.

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