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(C) Given that, $	an \alpha + 	an \beta + 	an \gamma = 	an \alpha 	an \beta 	an \gamma$.This implies $	an \alpha + 	an \beta = -	an \gamma (1

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$\pm 1$

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(C) Given that, $	an \alpha + 	an \beta + 	an \gamma = 	an \alpha 	an \beta 	an \gamma$.This implies $	an \alpha + 	an \beta = -	an \gamma (1 - 	an \alpha 	an \beta)$.Dividing by $(1 - 	an \alpha 	an \beta)$, we get $\frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = 	an(-\gamma)$.Thus, $	an(\alpha + \beta) = 	an(-\gamma)$, which means $\alpha + \beta = n\pi - \gamma$ for some integer $n \in \mathbb{Z}$.Therefore, $\alpha + \beta + \gamma = n\pi$.Using Euler's formula, $x = e^{i\alpha}$, $y = e^{i\beta}$, and $z = e^{i\gamma}$.Then $xyz = e^{i\alpha} \cdot e^{i\beta} \cdot e^{i\gamma} = e^{i(\alpha + \beta + \gamma)} = e^{in\pi}$.Since $e^{in\pi} = \cos(n\pi) + i\sin(n\pi) = \cos(n\pi)$, and $\cos(n\pi)$ is $1$ if $n$ is even and $-1$ if $n$ is odd.Thus, $xyz = \pm 1$.

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