Let $\Delta=\left|\begin{matrix}x+2&x+3&x+2a\\ x+3&x+4&x+2b\\ x+4&x+5&x+2c\end{matrix}\right|$
$=\frac{1}{2}\left|\begin{matrix}x+2&x+3&x+2a\\ 0&0&2\left(2b-a-c\right)\\ x+4&x+5&x+2c\end{matrix}\right|$
(using $R_{2} \to 2R_{2} - R_{1} - R_{3})$
But a, b and c are in AP using 2b = a + c, we get
$\Delta=\frac{1}{2}\left|\begin{matrix}x+2&x+3&x+2a\\ 0&0&0\\ x+4&x+5&x+2c\end{matrix}\right|=0$
Since, all elements of $R_{2}$ are zero.