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(A) Given that $\bar{a}$ and $\bar{c}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{c}| = 1$. The angle between them is $	heta = \frac{\pi}{3}$,so

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$10$

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(A) Given that $\bar{a}$ and $\bar{c}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{c}| = 1$. The angle between them is $	heta = \frac{\pi}{3}$,so $\bar{a} \cdot \bar{c} = |\bar{a}| |\bar{c}| \cos(\frac{\pi}{3}) = 1 	imes 1 	imes \frac{1}{2} = \frac{1}{2}$.Using the vector triple product formula,$\bar{a} 	imes (\bar{b} 	imes \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.Substitute this into the given equation: $((\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}) \cdot (\bar{a} 	imes \bar{c}) = 5$.Since $(\bar{a} 	imes \bar{c})$ is perpendicular to both $\bar{a}$ and $\bar{c}$,we have $\bar{c} \cdot (\bar{a} 	imes \bar{c}) = 0$.Thus,the equation simplifies to $(\bar{a} \cdot \bar{c}) \bar{b} \cdot (\bar{a} 	imes \bar{c}) = 5$.Substituting $\bar{a} \cdot \bar{c} = \frac{1}{2}$,we get $\frac{1}{2} [\bar{b} \bar{a} \bar{c}] = 5$,which implies $[\bar{b} \bar{a} \bar{c}] = 10$.Since the scalar triple product is cyclic,$[\bar{b} \bar{a} \bar{c}] = [\bar{a} \bar{b} \bar{c}] = 10$.

Let $\bar{a}$ and $\bar{c}$ be unit vectors at an angle $\frac{\pi}{3}$ with each other. If $(\bar{a} \times(\bar{b} \times \bar{c})) \cdot(\bar{a} \times \bar{c})=5$,then $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=$

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