3. (15 points) Fill seven different perfect squares into seven consecutive boxes so that the sum of any three adjacent boxes is greater than 100. Find the minimum possible value of the sum of these seven perfect squares.


3. Since the three perfect squares at positions $1, 4$, and $7$ are all different, considering the smallest possible values, one of these numbers must be at least 9. After positioning 9, the remaining six positions can be divided into two groups, each with three consecutive numbers whose sum is at least 101. Therefore, the sum of these seven numbers is at least $101 \times 2 + 9 = 211$. Thus, the smallest possible value is 211.

The arrangement $1, 36, 64, 9, 81, 16, 4$ satisfies the conditions, with a sum of 211.