Explanation

(a)
Given that, $\left| \begin{matrix}
x + 1 & x + 2 & x + 3 \\
x + 2 & x + 3 & x + 4 \\
x + a & x + b & x + c
\end{matrix} \right| = 0$
$\Rightarrow \ \left| \begin{matrix}
- 1 & - 1 & x + 3 \\
- 1 & - 1 & x + 4 \\
a - b & b - c & x + c
\end{matrix} \right| = 0$ $\ \binom{C_{1} \rightarrow C_{1} - C_{2}}{C_{2} \rightarrow C_{2} - C_{3}}$
$\Rightarrow \ \left| \begin{matrix}
0 & 0 & - 1 \\
- 1 & - 1 & x + 4 \\
a - b & b - c & x + c
\end{matrix} \right| = 0\ \ \ (R_{1} \rightarrow R_{1} - R_{2})$
$\Rightarrow ( - 1)( - b + c + a - b) = 0$
$\Rightarrow 2b - a - c = 0$
$\Rightarrow \ a + c = 2b$
$\therefore\ a,b,c$ in AP.