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(C) Given $|\overline{a}| = \sqrt{3}, |\overline{b}| = 1, |\overline{c}| = 2$. Using the vector triple product formula $\overline{a} 	imes (\overline

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$\frac{4}{3}$

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(C) Given $|\overline{a}| = \sqrt{3}, |\overline{b}| = 1, |\overline{c}| = 2$. Using the vector triple product formula $\overline{a} 	imes (\overline{a} 	imes \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{a} - (\overline{a} \cdot \overline{a}) \overline{c}$. Substituting this into the given equation: $(\overline{a} \cdot \overline{c}) \overline{a} - |\overline{a}|^2 \overline{c} + 3 \overline{b} = \overline{0}$. Since $|\overline{a}|^2 = (\sqrt{3})^2 = 3$,we have $(\overline{a} \cdot \overline{c}) \overline{a} - 3 \overline{c} = -3 \overline{b}$. Taking the squared magnitude on both sides: $|(\overline{a} \cdot \overline{c}) \overline{a} - 3 \overline{c}|^2 = |-3 \overline{b}|^2$. Expanding the left side: $(\overline{a} \cdot \overline{c})^2 |\overline{a}|^2 + 9 |\overline{c}|^2 - 6 (\overline{a} \cdot \overline{c})(\overline{a} \cdot \overline{c}) = 9 |\overline{b}|^2$. $(\overline{a} \cdot \overline{c})^2 (3) + 9 (2)^2 - 6 (\overline{a} \cdot \overline{c})^2 = 9 (1)^2$. $-3 (\overline{a} \cdot \overline{c})^2 + 36 = 9$. $-3 (\overline{a} \cdot \overline{c})^2 = -27 \implies (\overline{a} \cdot \overline{c})^2 = 9$. Thus,$\overline{a} \cdot \overline{c} = \pm 3$. Since $\overline{a} \cdot \overline{c} = |\overline{a}| |\overline{c}| \cos 	heta$,we have $(\sqrt{3})(2) \cos 	heta = \pm 3$. $\cos 	heta = \pm \frac{3}{2\sqrt{3}} = \pm \frac{\sqrt{3}}{2}$. Therefore,$\sec^2 	heta = \frac{1}{\cos^2 	heta} = \frac{1}{3/4} = \frac{4}{3}$.

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