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(C) Let $\vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$.Given $\vec{a} 	imes \vec{c} = \vec{c} 	imes \vec{b}$.This can be rewritten as $\vec{a} 	imes

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$4$

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(C) Let $\vec{c} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$.Given $\vec{a} 	imes \vec{c} = \vec{c} 	imes \vec{b}$.This can be rewritten as $\vec{a} 	imes \vec{c} + \vec{b} 	imes \vec{c} = \vec{0}$,which implies $(\vec{a} + \vec{b}) 	imes \vec{c} = \vec{0}$.This means the vector $(\vec{a} + \vec{b})$ is parallel to $\vec{c}$.Let $\vec{a} + \vec{b} = \lambda \vec{c}$ for some scalar $\lambda$.Taking the magnitude on both sides,$|\vec{a} + \vec{b}| = |\lambda| |\vec{c}|$.We know $|\vec{a} + \vec{b}| = \sqrt{29}$ and $|\vec{c}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.So,$\sqrt{29} = |\lambda| \sqrt{29}$,which gives $|\lambda| = 1$,so $\lambda = \pm 1$.Thus,$\vec{a} + \vec{b} = \pm(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$.Now,calculate the dot product: $(\vec{a} + \vec{b}) \cdot (-7 \hat{i} + 2 \hat{j} + 3 \hat{k}) = \pm(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (-7 \hat{i} + 2 \hat{j} + 3 \hat{k})$.$= \pm(2(-7) + 3(2) + 4(3)) = \pm(-14 + 6 + 12) = \pm(4)$.Thus,the possible values are $4$ or $-4$. Given the options,$4$ is a possible value.

If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$,then a possible value of $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

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