Step 1: Understand the goal.
We have a determinant $D$ built from factorials, and we must find $\dfrac{D}{n!\,(n+1)!\,(n+2)!}$. The plan is to pull common factors out of each row to make the numbers simple.
Step 2: Factor out from each row.
From row 1 take out $n!$, from row 2 take out $(n+1)!$, from row 3 take out $(n+2)!$. Using $(n+1)!=(n+1)\,n!$ and similar, the determinant becomes \[ \frac{D}{n!(n+1)!(n+2)!}=\begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 1 & n+2 & (n+2)(n+3)\\ 1 & n+3 & (n+3)(n+4) \end{vmatrix}. \]
Step 3: Make zeros using row subtraction.
Subtracting rows removes clutter. Do $R_2\to R_2-R_1$ and $R_3\to R_3-R_2$ to create zeros in the first column below the top.
Step 4: Simplify the entries.
The middle column differences give $1$ and $1$. The last column differences give $2(n+2)$ and $2(n+3)$. The determinant becomes \[ \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 0 & 1 & 2(n+2)\\ 0 & 1 & 2(n+3) \end{vmatrix}. \]
Step 5: One more subtraction.
Do $R_3\to R_3-R_2$ to clear the second column: \[ \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 0 & 1 & 2(n+2)\\ 0 & 0 & 2 \end{vmatrix}. \] This is upper triangular.
Step 6: Read the answer.
For an upper triangular determinant we just multiply the diagonal: $1\times1\times2=2$. \[ \boxed{2} \]