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(D) Given: $(\overline{a} 	imes \overline{b}) 	imes \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.Using the vector triple pro

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$\frac{3}{2 \sqrt{2}}$

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(D) Given: $(\overline{a} 	imes \overline{b}) 	imes \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.Using the vector triple product formula: $(\overline{a} 	imes \overline{b}) 	imes \overline{c} = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a}$.Since no two vectors are collinear,$\overline{a}$ and $\overline{b}$ are linearly independent. Thus,the coefficient of $\overline{b}$ must be zero,so $(\overline{a} \cdot \overline{c}) = 0$.Comparing the coefficients of $\overline{a}$,we get: $-(\overline{b} \cdot \overline{c}) = \frac{1}{3}|\overline{b}||\overline{c}|$.Substituting the definition of the dot product,$-|\overline{b}||\overline{c}| \cos 	heta = \frac{1}{3}|\overline{b}||\overline{c}|$.Since the vectors are non-zero,we divide by $|\overline{b}||\overline{c}|$: $\cos 	heta = -\frac{1}{3}$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we have $\sin^2 	heta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.Thus,$\sin 	heta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Therefore,$\operatorname{cosec} 	heta = \frac{1}{\sin 	heta} = \frac{3}{2\sqrt{2}}$.

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