Given that $\begin{vmatrix}
\alpha p + \beta & \alpha & \beta \\
\beta p + \gamma & \beta & \gamma \\
0 & \alpha p + \beta & \beta p + \gamma
\end{vmatrix} = 0$
Expand along with $C_{1},$ we get
$(\alpha p + \beta) \begin{vmatrix}
\beta & \gamma \\
\alpha p + \beta & \beta p + \gamma
\end{vmatrix} - (\beta p + \gamma) \begin{vmatrix}
\alpha & \beta \\
\alpha p + \beta & \beta p + \gamma
\end{vmatrix} = 0$
Perform the operation $R_{2} \rightarrow R_{2} - R_{1},$ in the first determinant and in the second determinant $R_{2} \rightarrow R_{2} - p R_{1},$ we get
$(\alpha p + \beta) \begin{vmatrix}
\beta & \gamma \\
\alpha p & \beta p
\end{vmatrix} - (\beta p + \gamma) \begin{vmatrix}
\alpha & \beta \\
\beta & \gamma
\end{vmatrix} = 0$
$\implies (\alpha p + \beta) \begin{vmatrix}
\beta & \gamma \\
\alpha p & \beta p
\end{vmatrix} = (\beta p + \gamma) \begin{vmatrix}
\alpha & \beta \\
\beta & \gamma
\end{vmatrix}$
$\implies p(\alpha p + \beta) \begin{vmatrix}
\beta & \gamma \\
\alpha & \beta
\end{vmatrix} = (\beta p + \gamma) \begin{vmatrix}
\alpha & \beta \\
\beta & \gamma
\end{vmatrix}$
$\implies p(\alpha p + \beta)(\beta^{2} - \alpha \gamma) = (\beta p + \gamma) (\alpha \gamma - \beta^{2})$
$\implies p(\alpha p + \beta)(\beta^{2} - \alpha \gamma) = -(\beta p + \gamma) (\beta^{2} - \alpha \gamma)$
$\implies p(\alpha p + \beta)(\beta^{2} - \alpha \gamma) + (\beta p + \gamma) (\beta^{2} - \alpha \gamma) = 0$
$\implies (\beta^{2} - \alpha \gamma) [p(\alpha p + \beta) + (\beta p + \gamma)] = 0$
From above equation, either $(\beta^{2} -\alpha \gamma) = 0$ or $p(\alpha p + \beta) + (\beta p + \gamma) = 0$
$\implies \beta^{2} - \alpha \gamma = 0$
$\implies \beta^{2} = \alpha \gamma$
$\therefore \alpha,\beta,\gamma$ are in G.P.
So, the correct answer is $(A).$
