First, let's determine the number of elements in the Cartesian product $$A \times B$$. If set $$A$$ has 5 elements and set $$B$$ has 2 elements, then the number of elements in $$A \times B$$ is:

$$|A \times B| = |A| \times |B| = 5 \times 2 = 10$$

We need to find the number of subsets of $$A \times B$$ with at least 3 elements and at most 6 elements. The total number of elements in $$A \times B$$ is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements.

The number of ways to choose $$r$$ elements from a set of 10 elements is given by the binomial coefficient:

$$\binom{10}{r} = \frac{10!}{r!(10-r)!}$$

We will find the sum of the binomial coefficients for $$r = 3, 4, 5, 6$$:

First, $$\binom{10}{3}$$:

$$\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$

Next, $$\binom{10}{4}$$:

$$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$

Next, $$\binom{10}{5}$$:

$$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$$

Next, $$\binom{10}{6}$$:

$$\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$

Finally, we sum these values:

$$\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792$$

Therefore, the number of subsets of $$A \times B$$ each having at least 3 and at most 6 elements is:

Option D - 792