If $\alpha,\beta,\gamma$ are such that $\alpha+\beta+\gamma=0$ then $\begin{vmatrix}1&\cos \gamma&\cos \beta\\\cos \alpha& 1&\cos\alpha\\\cos \beta&\cos \alpha&1\end{vmatrix}$ is equal to

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

Let $A,B,C$ be numbers such that
$\alpha=B-C$
$\beta=C-A$
$\gamma=A-B$
So that $\alpha+\beta+\gamma=0$
$\therefore$ Given determinant $\begin{vmatrix} 1&\cos(A-B) &\cos(C-A)\\\cos(A-B)& 1&\cos(B-C)\\\cos(C-A)&\cos(B-C) &1\end{vmatrix}$
$\Rightarrow \begin{vmatrix} \cos^2A+\sin^2A&\cos A\cos B+\sin A\sin B&\cos C\cos A+\sin A\sin C\\\cos A\cos B+\sin A\sin B&\cos^2B+\sin ^2B&\cos B\cos C+\sin B\sin C\\\cos C\cos A+\sin C\sin A&\cos B\cos C+\sin B\sin C&\cos^2C+\sin^2C\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\cos A&\sin A&0\\\cos B&\sin B&0\\\cos C&\sin C&0\end{vmatrix}\begin{vmatrix}\cos A&\sin A&0\\\cos B&\sin B&0\\\cos C&\sin C&0\end{vmatrix}=0$
Hence (D) is the correct answer.