If $\alpha, \beta$ and $\gamma$ are such that $\alpha+\beta+\gamma=0$, then the value of $\left|\begin{array}{ccc}1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$ is
Answer & explanation
Correct answer: option 1
Operating $C_2 \rightarrow C_2-\cos \gamma~ C_1, C_3 \rightarrow C_3-\cos \beta C_1$,
$\Delta=\left|\begin{array}{ccc}1 & 0 & 0 \\ \cos \gamma & \sin ^2 \gamma & \cos \alpha-\cos \beta \cos \gamma \\ \cos \beta & \cos \alpha-\cos \beta \cos \gamma & \sin ^2 \beta \end{array}\right|$
$\Delta=\left|\begin{array}{ccc}1 & 0 & 0 \\ \cos \gamma & \sin ^2 \gamma & -\sin \beta \sin \gamma \\ \cos \beta & -\sin \gamma \sin \beta & \sin ^2 \beta\end{array}\right|\left\{\begin{array}{c}∵ \alpha=-(\beta+\gamma) \\ ∴ \cos \alpha=(\beta+\gamma) \\ \cos \alpha=\cos \beta \cos \gamma-\sin \beta \sin \gamma\end{array}\right\}$
$=1\left(\sin ^2 \gamma \sin ^2 \beta-\sin ^2 \gamma \sin ^2 \beta\right)=0$
Hence (1) is the correct answer.