We can approach this problem using combinations.

First, we choose 2 teachers to be assigned to School 1. This can be done in C(12,2) ways.

Next, we need to assign the remaining 10 teachers to the three schools. One possible way to do this is to use a stars and bars approach. Imagine lining up the 10 teachers in a row, and placing 2 separators (represented by bars) to divide them into 3 groups of teachers for each school. For example:

|T T T T T|T T|T T T

This represents 5 teachers for School 1, 2 teachers for School 2, and 3 teachers for School 3.

There are 9 spaces between the teachers (including the ends) where we can place the 2 bars. This can be done in C(9,2) ways.

Therefore, the total number of ways to assign 12 teachers to 3 schools such that School 1 receives at least 2 teachers is:

C(12,2) x C(9,2) = 66 x 36 = 2376 ways.