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(D) Given that $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ are unit vectors, so $|\hat{\alpha}| = |\hat{\beta}| = |\hat{\gamma}| = 1$.Using the vector t

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$\frac{2 \pi}{3}$

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(D) Given that $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ are unit vectors, so $|\hat{\alpha}| = |\hat{\beta}| = |\hat{\gamma}| = 1$.Using the vector triple product formula: $\hat{\alpha} 	imes (\hat{\beta} 	imes \hat{\gamma}) = (\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta} - (\hat{\alpha} \cdot \hat{\beta}) \hat{\gamma}$.Given the equation: $(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta} - (\hat{\alpha} \cdot \hat{\beta}) \hat{\gamma} = \frac{1}{2} \hat{\beta} + \frac{1}{2} \hat{\gamma}$.Since $\hat{\beta}$ and $\hat{\gamma}$ are not parallel, we can compare the coefficients of $\hat{\beta}$ and $\hat{\gamma}$ on both sides.Comparing the coefficients of $\hat{\gamma}$, we get: $-(\hat{\alpha} \cdot \hat{\beta}) = \frac{1}{2}$.Thus, $\hat{\alpha} \cdot \hat{\beta} = -\frac{1}{2}$.Since $\hat{\alpha} \cdot \hat{\beta} = |\hat{\alpha}| |\hat{\beta}| \cos 	heta$, where $	heta$ is the angle between $\hat{\alpha}$ and $\hat{\beta}$, we have $1 	imes 1 	imes \cos 	heta = -\frac{1}{2}$.Therefore, $\cos 	heta = -\frac{1}{2}$, which implies $	heta = \frac{2 \pi}{3}$.

Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma}) = \frac{1}{2}(\hat{\beta} + \hat{\gamma})$. If $\hat{\beta}$ is not parallel to $\hat{\gamma}$, then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is

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