Consider a sequence of eleven squares that have side lengths $3, 6, 9, 12,\ldots, 33$. Eleven copies of a single square each with area $A$ have the same total area as the total area of the eleven squares of the sequence. Find $A$.
1. **Identify the sequence of side lengths and their areas:**
   The side lengths of the squares are given by the arithmetic sequence: \(3, 6, 9, 12, \ldots, 33\). The general term for the \(n\)-th side length is:
   \[
   a_n = 3n \quad \text{for} \quad n = 1, 2, \ldots, 11
   \]
   The area of each square is:
   \[
   a_n^2 = (3n)^2 = 9n^2
   \]

2. **Sum the areas of the squares:**
   We need to find the sum of the areas of these 11 squares:
   \[
   \sum_{n=1}^{11} 9n^2
   \]
   Factor out the constant 9:
   \[
   9 \sum_{n=1}^{11} n^2
   \]

3. **Use the formula for the sum of squares:**
   The sum of the squares of the first \(m\) natural numbers is given by:
   \[
   \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}
   \]
   For \(m = 11\):
   \[
   \sum_{n=1}^{11} n^2 = \frac{11 \cdot 12 \cdot 23}{6}
   \]

4. **Calculate the sum:**
   \[
   \sum_{n=1}^{11} n^2 = \frac{11 \cdot 12 \cdot 23}{6} = \frac{3036}{6} = 506
   \]

5. **Multiply by 9 to get the total area:**
   \[
   9 \sum_{n=1}^{11} n^2 = 9 \cdot 506 = 4554
   \]

6. **Set up the equation for the area \(A\):**
   The total area of the eleven squares is equal to 11 times the area \(A\) of the single square:
   \[
   11A = 4554
   \]

7. **Solve for \(A\):**
   \[
   A = \frac{4554}{11} = 414
   \]

The final answer is \(\boxed{414}\)