As given $\left|\begin{array}{lll}x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c\end{array}\right|=0$
$=\left|\begin{array}{ccc}-1 & -1 & x+3 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c\end{array}\right|=0 \quad \begin{aligned} & C_1 \rightarrow C_1-C_2 \\ & C_2 \rightarrow C_2-C_3\end{aligned}$
$\Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c\end{array}\right|=0$, by $R_1 \rightarrow R_1-R_2$
$\Rightarrow(-1)(-b+c+a-b)=0$
$\Rightarrow \quad 2 b-a-c=0 \Rightarrow a+c=2 b$ i.e., $a, b, c$ are in A.P.
Trick : In such type of problem, put any suitable value of $x$ i.e. 0 , so that the determinant.
$\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 4 \\ a & b & c\end{array}\right|=0$
$\begin{aligned} & \Rightarrow 1(3 c-4 b)-2(2 c-4 a)+3(2 b-3 a)=0 \\ & \Rightarrow-c+2 b-a=0 \Rightarrow 2 b=a+c \text {. Hence the result. }\end{aligned}$