Show that

Given: $\alpha, \beta, \gamma$ are in A.P.

Now,

$2 \beta=\alpha+\gamma$

$\Delta=\left|\begin{array}{lll}x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}x-3 & x-4 & x-\alpha \\ 2 x-4 & 2 x-6 & 2 x-2 \beta \\ x-1 & x-2 & x-\gamma\end{array}\right|$ [Applying $R_{2} \rightarrow 2 R_{2}$ ]

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}x-3 & x-4 & x-\alpha \\ 0 & 0 & -2 \beta+\alpha+\gamma \\ x-1 & x-2 & x-\gamma\end{array}\right|$

$[\because 2 \beta=\alpha+\gamma] \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-\left(R_{1}+R_{3}\right)\right]$

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}x-3 & x-4 & x-\alpha \\ 0 & 0 & 0 \\ x-1 & x-2 & x-\gamma\end{array}\right|$

$\Delta=0$