Given $a, b, c$ are in A.P.
$\therefore \quad 2 \mathrm{~b}=\mathrm{a}+\mathrm{c} ......(i)$
Now, $\left|\begin{array}{ccc}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|$
[Applying $\left.R_{2} \rightarrow 2 R_{2}\right]$
$$
\begin{array}{c}
=\frac{1}{2}\left|\begin{array}{ccc}
\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+\mathrm{a} \\
2 \mathrm{x}+4 & 2 \mathrm{x}+6 & 2 \mathrm{x}+2 \mathrm{~b} \\
\mathrm{x}+3 & \mathrm{x}+4 & \mathrm{x}+\mathrm{c}
\end{array}\right| \\
=\frac{1}{2}\left|\begin{array}{ccc}
\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+\mathrm{a} \\
2 \mathrm{x}+4 & 2 \mathrm{x}+6 & 2 \mathrm{x}+(\mathrm{a}+\mathrm{c}) \\
\mathrm{x}+3 & \mathrm{x}+4 & \mathrm{x}+\mathrm{c}
\end{array}\right|
\end{array}
$$
[using equation (i)]
$$
=\frac{1}{2}\left|\begin{array}{ccc}
x+1 & x+2 & x+a \\
0 & 0 & 0 \\
x+3 & x+4 & x+c
\end{array}\right|=\frac{1}{2} \cdot 0=0
$$
$\left[\right.$ Applying $\left.\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\left(\mathrm{R}_{1}+\mathrm{R}_{3}\right)\right]$