If $\left|\begin{array}{ccc}x+\alpha & \beta & \gamma \\ \alpha & x+\beta & \gamma \\ \alpha & \beta & x+\gamma\end{array}\right|=0$, then x is equal to

$0,-(\alpha+\beta+\gamma)$

applying C₁ → C₁ + C₂ + C₃

$\Rightarrow(x+\alpha+\beta+\gamma)\left|\begin{array}{ccc}1 & \beta & \gamma \\ 1 & x+\beta & \gamma \\ 1 & \beta & x+\gamma\end{array}\right|=0$

applying $R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3-R_1$

$\Rightarrow(x+\alpha+\beta+\gamma)\left|\begin{array}{ccc}1 & \beta & \gamma \\ 0 & x & 0 \\ 0 & 0 & x\end{array}\right|=0$

$\Rightarrow(x+\alpha+\beta+\gamma) . x^2=0 \Rightarrow x=0,-(\alpha+\beta+\gamma)$

Hence (1) is the correct answer.