$=\left|\begin{array}{llc}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+(a+c) \\ x+4 & x+5 & x+2 c\end{array}\right|$ $(2 b=a+c \text { as } a, b, \text { and } c \text { are in } A P)$

Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{3} \rightarrow R_{3}-R_{2},$ we have:

$\Delta=\left|\begin{array}{ccc}-1 & -1 & a-c \\ x+3 & x+4 & x+(a+c) \\ 1 & 1 & c-a\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}0 & 0 & 0 \\ x+3 & x+4 & x+a+c \\ 1 & 1 & c-a\end{array}\right|$

Here, all the elements of the first row ( $\mathrm{R}_{1}$ ) are zero.

Hence, we have $\Delta=0$ The correct

answer is $D$.