Solution:

Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$
$\begin{array}{l}
=\left|\begin{array}{ccc}
(a+1)(a+2)-(a+2)(a+3) & a+2-a-3 & 0 \\
(a+2)(a+3)-(a+3)(a+4) & a+3-a-4 & 0 \\
(a+3)(a+4) & & a+4 & 1
\end{array}\right| \\
=\left|\begin{array}{ccc}
(a+2)(a+1-a-3) & -1 & 0 \\
(a+3)(a+2-a-4) & -1 & 0 \\
(a+3)(a+4) & a+4 & 1
\end{array}\right|
\end{array}$
$=\left|\begin{array}{ccc}
-2(a+2) & -1 & 0 \\
-2(a+3) & -1 & 0 \\
(a+3)(a+4) & a+4 & 1
\end{array}\right|$
Expanding with $C_{3}$
$\begin{array}{l}
=(2(a+2)-2(a+3)) \\
=(2 a+4-2 a-6) \\
=-2
\end{array}$