# If $\alpha,\beta,\gamma$ are real numbers then,$\begin{vmatrix} 1& \cos(\beta-\alpha)&\cos(\gamma-\alpha)\\\cos(\alpha-\beta)&1&\cos(\gamma-\beta)\\\cos(\alpha-\gamma)&\cos(\beta-\gamma)&1\end{vmatrix}$ is equal to

$\begin{array}{1 1}(A)\;-1&(B)\;\cos\alpha\cos\beta\cos\gamma\\(C)\;\cos \alpha+\cos\beta+\cos\gamma&(D)\;0\end{array}$

$\Delta=\begin{vmatrix} 1& \cos(\beta-\alpha)&\cos(\gamma-\alpha)\\\cos(\alpha-\beta)&1&\cos(\gamma-\beta)\\\cos(\alpha-\gamma)&\cos(\beta-\gamma)&1\end{vmatrix}$
By using the formula we have
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$\Delta=\begin{vmatrix}\cos \alpha&\sin \alpha&0\\\cos\beta&\sin\beta&0\\\cos\gamma&\sin \gamma&0\end{vmatrix}.\begin{vmatrix}\cos \alpha&\sin \alpha&0\\\cos\beta&\sin\beta&0\\\cos\gamma&\sin \gamma&0\end{vmatrix}$
$\Rightarrow 0$
Hence (D) is the correct answer.