Explanation

(d)
Given $\left| \begin{matrix}
x + 1 & x + 2 & x + a \\
x + 2 & x + 3 & x + b \\
x + 3 & x + 4 & x + c
\end{matrix} \right| = 0$
Applying $R_{1} \rightarrow + R_{1} + R_{3} - 2R_{2},\$we get
$\left| \begin{matrix}
0 & 0 & a + c - 2b \\
x + 2 & x + 3 & x + b \\
x + 3 & x + 4 & x + c
\end{matrix} \right| = 0$
$\Rightarrow (a + c - 2b)\left\lbrack x^{2} + 6x + 8 - \left( x^{2} + 6x + 9 \right) \right\rbrack = 0$
$\Rightarrow (a + c - 2b)( - 1) = 0$
$\Rightarrow 2b = a + c$
$\Rightarrow a,b,c\$are in AP