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(D) The given determinant is $D = \begin{vmatrix} 1 & \cos(\beta - \alpha) & \cos(\gamma - \alpha) \ \cos(\alpha - \beta) & 1 & \cos(\gamma - \beta)

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$0$

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(D) The given determinant is $D = \begin{vmatrix} 1 & \cos(\beta - \alpha) & \cos(\gamma - \alpha) \ \cos(\alpha - \beta) & 1 & \cos(\gamma - \beta) \ \cos(\alpha - \gamma) & \cos(\beta - \gamma) & 1 \end{vmatrix}$.Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$ and $1 = \cos^2 	heta + \sin^2 	heta$,we can write each element as a dot product of two vectors.Let $v_1 = (\cos \alpha, \sin \alpha)$,$v_2 = (\cos \beta, \sin \beta)$,and $v_3 = (\cos \gamma, \sin \gamma)$.Then the determinant $D$ can be written as the product of two matrices:$D = \begin{vmatrix} \cos \alpha & \sin \alpha & 0 \ \cos \beta & \sin \beta & 0 \ \cos \gamma & \sin \gamma & 0 \end{vmatrix} 	imes \begin{vmatrix} \cos \alpha & \cos \beta & \cos \gamma \ \sin \alpha & \sin \beta & \sin \gamma \ 0 & 0 & 0 \end{vmatrix}$.Since the third column of the first matrix is all zeros,the determinant of the first matrix is $0$.Therefore,$D = 0 	imes 0 = 0$.

If $\alpha, \beta, \text{ and } \gamma$ are real numbers,then $D = \begin{vmatrix} 1 & \cos(\beta - \alpha) & \cos(\gamma - \alpha) \\ \cos(\alpha - \beta) & 1 & \cos(\gamma - \beta) \\ \cos(\alpha - \gamma) & \cos(\beta - \gamma) & 1 \end{vmatrix} = $

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