Let $A=\frac{2^2+3\cdot 2 + 1}{3! \cdot 4!} + \frac{3^2+3\cdot 3 + 1}{4! \cdot 5!} + \frac{4^2+3\cdot 4 + 1}{5! \cdot 6!} + \dots + \frac{10^2+3\cdot 10 + 1}{11! \cdot 12!}$. What is the remainder when $11!\cdot 12! \cdot A$ is divided by $11$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 8
\qquad\textbf{(E)}\ 10
$
1. We start with the given expression for \( A \):
   \[
   A = \sum_{n=2}^{10} \frac{n^2 + 3n + 1}{(n+1)! \cdot (n+2)!}
   \]

2. We need to find the remainder when \( 11! \cdot 12! \cdot A \) is divided by 11. First, let's simplify the expression for \( A \):
   \[
   A = \sum_{n=2}^{10} \frac{n^2 + 3n + 1}{(n+1)! \cdot (n+2)!}
   \]

3. Notice that:
   \[
   (n+1)! \cdot (n+2)! = (n+1)! \cdot (n+2) \cdot (n+1)!
   \]
   \[
   = (n+1)!^2 \cdot (n+2)
   \]

4. Therefore, the term inside the sum can be rewritten as:
   \[
   \frac{n^2 + 3n + 1}{(n+1)! \cdot (n+2)!} = \frac{n^2 + 3n + 1}{(n+1)!^2 \cdot (n+2)}
   \]

5. Now, we multiply \( A \) by \( 11! \cdot 12! \):
   \[
   11! \cdot 12! \cdot A = 11! \cdot 12! \cdot \sum_{n=2}^{10} \frac{n^2 + 3n + 1}{(n+1)!^2 \cdot (n+2)}
   \]

6. We can factor out \( 11! \cdot 12! \) and distribute it inside the sum:
   \[
   11! \cdot 12! \cdot A = \sum_{n=2}^{10} \frac{11! \cdot 12! \cdot (n^2 + 3n + 1)}{(n+1)!^2 \cdot (n+2)}
   \]

7. For each term in the sum, we can simplify the factorials. Notice that for \( n \leq 10 \), \( 11! \) and \( 12! \) will cancel out the factorials in the denominator:
   \[
   \frac{11! \cdot 12!}{(n+1)!^2 \cdot (n+2)} = \frac{11! \cdot 12!}{(n+1)! \cdot (n+1)! \cdot (n+2)}
   \]

8. For the last term when \( n = 10 \):
   \[
   \frac{11! \cdot 12! \cdot (10^2 + 3 \cdot 10 + 1)}{11! \cdot 12!} = 10^2 + 3 \cdot 10 + 1 = 100 + 30 + 1 = 131
   \]

9. We need to find the remainder of \( 131 \) when divided by 11:
   \[
   131 \div 11 = 11 \text{ remainder } 10
   \]

10. Therefore, the remainder when \( 11! \cdot 12! \cdot A \) is divided by 11 is 10.

The final answer is \(\boxed{10}\)