Let $\Delta=\begin{vmatrix}x+1 & x+2 & x+a\\x+2 & x+3 & x+b\\x+3 & x+4 & x+c\end{vmatrix}$
Apply $R_1\rightarrow R_2-R_1$ and $R_2\rightarrow R_3-R_2$
$\Delta=\begin{vmatrix}1 & 1 & b-a\\1 & 1 & c-b\\x+3 & x+4 & x+c\end{vmatrix}$
Apply $C_1\rightarrow C_1-C_2$
$\Delta=\begin{vmatrix}0 & 1 & b-a\\0 & 1 & c-b\\-1 & x+4 & x+c\end{vmatrix}$
Now expanding along $C_1$ we get,
$=0+0-1((b-a)-(c-b))=-b+a+c-b=a+c-2b$
But since a,b,c are in A.P
2b=a+c
Therefore $\Delta=2b-2b=0.$
Hence the value is 0.
So the statement is True.