To find the middle term in the expansion of (2x + 3y)^12, we need to use the Binomial Theorem. The Binomial Theorem states that for any positive integer n:

(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n

Where C(n,r) represents the binomial coefficient or combination of "n choose r", given by:

C(n,r) = n! / (r! * (n-r)!)

Now let's apply this formula to find the middle term in the expansion of (2x + 3y)^12.

Step 1: Determine the middle term
The middle term in the expansion of (2x + 3y)^12 will have an equal number of terms before and after it. Since there are 13 terms in the expansion (starting from n = 0 to n = 12), the middle term will be at n = 6.

Step 2: Find the middle term coefficient
To find the coefficient of the middle term, we need to use the binomial coefficient C(12,6). Plugging the values into the formula, we have:

C(12,6) = 12! / (6! * (12-6)!)
= 12! / (6! * 6!)
= (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1)
= 924

Step 3: Write the middle term
The middle term will have the form:

C(12,6) * (2x)^(12-6) * (3y)^6

Plugging in the values, we get:

924 * (2x)^6 * (3y)^6

Simplifying this expression gives us the middle term:

(924 * 2^6 * 3^6) * x^6 * y^6

So, the correct answer is option B) 12C6 * (2x)^6 * (3y)^6.