Let vec{a} = 4hat{i} - hat{j} + 3hat{k}, vec{ | vedclass

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(D) Given $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$.This can be written as $(2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$.Th

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-$3$

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(D) Given $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$.This can be written as $(2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$.This implies that $\vec{c}$ is parallel to $(2\vec{a} + 3\vec{b})$, so $\vec{c} = k(2\vec{a} + 3\vec{b})$ for some scalar $k$.Calculate $2\vec{a} + 3\vec{b} = 2(4\hat{i} - \hat{j} + 3\hat{k}) + 3(10\hat{i} + 2\hat{j} - \hat{k}) = (8+30)\hat{i} + (-2+6)\hat{j} + (6-3)\hat{k} = 38\hat{i} + 4\hat{j} + 3\hat{k}$.Thus, $\vec{c} = k(38\hat{i} + 4\hat{j} + 3\hat{k})$.Given $\vec{a} \cdot \vec{c} = 15$, substitute $\vec{a}$ and $\vec{c}$:$(4\hat{i} - \hat{j} + 3\hat{k}) \cdot k(38\hat{i} + 4\hat{j} + 3\hat{k}) = 15$.$k(4 	imes 38 + (-1) 	imes 4 + 3 	imes 3) = 15$.$k(152 - 4 + 9) = 15 \Rightarrow 157k = 15 \Rightarrow k = 15/157$.Now, calculate $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k}) = k(38\hat{i} + 4\hat{j} + 3\hat{k}) \cdot (\hat{i} + \hat{j} - 3\hat{k})$.$= k(38 	imes 1 + 4 	imes 1 + 3 	imes (-3)) = k(38 + 4 - 9) = 33k$.Substituting $k = 15/157$, we get $33 	imes (15/157) = 495/157 \approx 3.15$. Note: Re-evaluating the options provided, there may be a typo in the question constants. Based on the calculation, the result is $495/157$.

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to:

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