Explanation

(d)
Let $\mathrm{\Delta} = \left| \begin{matrix}
x + 2 & x + 3 & x + a \\
x + 4 & x + 5 & x + b \\
x + 6 & x + 7 & x + c
\end{matrix} \right|$
Applying $C_{2} \rightarrow C_{2} - C_{1}$ ,we get
$\mathrm{\Delta} = \left| \begin{matrix}
x + 2 & 1 & x + a \\
x + 4 & 1 & x + b \\
x + 6 & 1 & x + c
\end{matrix} \right|$
Applying $R_{2} \rightarrow R_{2} - R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$
$\Rightarrow \ \mathrm{\Delta} = \left| \begin{matrix}
x + 2 & 1 & x + a \\
2 & 0 & b - a \\
4 & 0 & c - a
\end{matrix} \right|$
$= - 1(2c - 2a - 4b + 4a)$
$\Rightarrow \ \mathrm{\Delta} = 2(2b - c - a)$ …(i)
Since, $a,b,c$ are in AP.
$\therefore b = \frac{a + c}{2}$
$\therefore\ \mathrm{\Delta} = 2(a + c - c - a)$
$= 0$ [from Eq. (i)]