Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in $$A \cap B$$, which are divisible by 3, is :

Set $$A$$ consists of the first 101 terms of the AP with first term 1 and common difference 5, so the general term is $$a_n = 5n - 4$$ for $$n = 1, 2, \ldots, 101$$. The last term is $$5(101) - 4 = 501$$.

Set $$B$$ consists of the first 71 terms of the AP with first term 9 and common difference 7, so the general term is $$b_m = 7m + 2$$ for $$m = 1, 2, \ldots, 71$$. The last term is $$7(71) + 2 = 499$$.

For a number to be in $$A \cap B$$, we need $$5n - 4 = 7m + 2$$, i.e., $$5n = 7m + 6$$. Taking modulo 7: $$5n \equiv 6 \pmod{7}$$. Since $$5 \times 3 = 15 \equiv 1 \pmod{7}$$, we get $$n \equiv 18 \equiv 4 \pmod{7}$$.

So the common terms form an AP: when $$n = 4$$, the term is $$5(4) - 4 = 16$$. The common difference is $$\text{lcm}(5, 7) = 35$$. The common terms are $$16, 51, 86, 121, \ldots$$, which can be written as $$16 + 35k$$ for $$k = 0, 1, 2, \ldots$$

We need $$16 + 35k \leq 499$$, so $$k \leq \frac{483}{35} = 13.8$$. Thus $$k = 0, 1, 2, \ldots, 13$$, giving 14 common terms.

Now we find which of these are divisible by 3. We need $$16 + 35k \equiv 0 \pmod{3}$$. Since $$16 \equiv 1 \pmod{3}$$ and $$35 \equiv 2 \pmod{3}$$:

$$1 + 2k \equiv 0 \pmod{3} \implies 2k \equiv 2 \pmod{3} \implies k \equiv 1 \pmod{3}$$

So $$k = 1, 4, 7, 10, 13$$, giving the terms $$51, 156, 261, 366, 471$$. That is 5 elements.

Hence, the correct answer is Option 2.