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(D) Let the determinant be $\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \ \sin \beta & \cos \beta & \sin(\beta +

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(D) Let the determinant be $\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \ \sin \beta & \cos \beta & \sin(\beta + \gamma) \ \sin \delta & \cos \delta & \sin(\delta + \gamma) \end{matrix} ight|$.Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can expand the third column:$\sin(\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma$$\sin(\beta + \gamma) = \sin \beta \cos \gamma + \cos \beta \sin \gamma$$\sin(\delta + \gamma) = \sin \delta \cos \gamma + \cos \delta \sin \gamma$Now,the determinant becomes:$\Delta = \left| \begin{matrix} \sin \alpha & \cos \alpha & \sin \alpha \cos \gamma + \cos \alpha \sin \gamma \ \sin \beta & \cos \beta & \sin \beta \cos \gamma + \cos \beta \sin \gamma \ \sin \delta & \cos \delta & \sin \delta \cos \gamma + \cos \delta \sin \gamma \end{matrix} ight|$.Apply the column operation $C_3 ightarrow C_3 - (\cos \gamma) C_1 - (\sin \gamma) C_2$:Each element in the third column becomes $(\sin 	heta \cos \gamma + \cos 	heta \sin \gamma) - (\cos \gamma)(\sin 	heta) - (\sin \gamma)(\cos 	heta) = 0$.Since all elements of the third column are $0$,the value of the determinant is $\Delta = 0$.

The value of $\left| \begin{matrix} \sin \alpha & \cos \alpha & \sin(\alpha + \gamma) \\ \sin \beta & \cos \beta & \sin(\beta + \gamma) \\ \sin \delta & \cos \delta & \sin(\delta + \gamma) \end{matrix} \right|$ is

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