If |vec{c}|^2 = 60 and vec{c} times (hat{i} + | vedclass

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(D) Let $\vec{c} = a\hat{i} + b\hat{j} + c\hat{k}$.Given $\vec{c} 	imes (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,this implies that $\vec{c}$ is par

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$12\sqrt{2}$

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(D) Let $\vec{c} = a\hat{i} + b\hat{j} + c\hat{k}$.Given $\vec{c} 	imes (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,this implies that $\vec{c}$ is parallel to the vector $\vec{v} = \hat{i} + 2\hat{j} + 5\hat{k}$.Thus,$\vec{c} = k(\hat{i} + 2\hat{j} + 5\hat{k})$ for some scalar $k$.Given $|\vec{c}|^2 = 60$,we have $k^2(1^2 + 2^2 + 5^2) = 60$.$k^2(1 + 4 + 25) = 60 \Rightarrow 30k^2 = 60 \Rightarrow k^2 = 2 \Rightarrow k = \pm\sqrt{2}$.Taking $k = \sqrt{2}$,we get $\vec{c} = \sqrt{2}\hat{i} + 2\sqrt{2}\hat{j} + 5\sqrt{2}\hat{k}$.Now,calculate the dot product $\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$:$= (\sqrt{2}\hat{i} + 2\sqrt{2}\hat{j} + 5\sqrt{2}\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$$= \sqrt{2}(-7) + 2\sqrt{2}(2) + 5\sqrt{2}(3)$$= -7\sqrt{2} + 4\sqrt{2} + 15\sqrt{2} = 12\sqrt{2}$.

If $|\vec{c}|^2 = 60$ and $\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$,then a value of $\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ is

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