If for positive integers $$r > 1$$, $$n > 2$$, the coefficients of the $$(3r)^{th}$$ and $$(r+2)^{th}$$ powers of $$x$$ in the expansion of $$(1 + x)^{2n}$$ are equal, then $$n$$ is equal to:

The expansion of $$(1 + x)^{2n}$$ is given by the binomial theorem, and the general term is $$T_{k+1} = \binom{2n}{k} x^k$$, where the coefficient of $$x^k$$ is $$\binom{2n}{k}$$. According to the problem, the coefficients of $$x^{3r}$$ and $$x^{r+2}$$ are equal, so we have:

$$\binom{2n}{3r} = \binom{2n}{r+2}$$

We know that $$\binom{2n}{a} = \binom{2n}{b}$$ implies either $$a = b$$ or $$a + b = 2n$$. Therefore, we consider two cases.

Case 1: $$3r = r + 2$$

Solving this equation:

$$3r - r = 2$$

$$2r = 2$$

$$r = 1$$

But the problem specifies that $$r > 1$$, so $$r = 1$$ is invalid. This case is discarded.

Case 2: $$3r + (r + 2) = 2n$$

Solving this equation:

$$3r + r + 2 = 2n$$

$$4r + 2 = 2n$$

Divide both sides by 2:

$$2r + 1 = n$$

So, $$n = 2r + 1$$.

Now, we verify the conditions. Given $$r > 1$$, $$n = 2r + 1 > 2(1) + 1 = 3 > 2$$, so $$n > 2$$ is satisfied. Also, the indices $$3r$$ and $$r+2$$ must be within the range $$[0, 2n]$$. With $$n = 2r + 1$$, $$2n = 4r + 2$$.

Check $$3r$$: since $$r > 1$$, $$3r > 0$$ and $$3r \leq 4r + 2$$ (as $$3r \leq 4r + 2$$ simplifies to $$0 \leq r + 2$$, true for $$r > 1$$).

Check $$r+2$$: $$r+2 \leq 4r + 2$$ simplifies to $$0 \leq 3r$$, true for $$r > 1$$.

Additionally, $$3r \neq r+2$$ for $$r > 1$$ (since $$3r = r+2$$ gives $$r=1$$, excluded), so Case 2 applies.

Now, comparing with the options:

A. $$2r + 1$$

B. $$2r - 1$$

C. $$3r$$

D. $$r + 1$$

Option A matches $$n = 2r + 1$$.

To verify, let $$r = 2$$ (since $$r > 1$$), then $$n = 2(2) + 1 = 5$$, so $$2n = 10$$. Coefficient of $$x^{3r} = x^6$$ is $$\binom{10}{6}$$, and coefficient of $$x^{r+2} = x^4$$ is $$\binom{10}{4}$$. Since $$\binom{10}{6} = \binom{10}{4}$$ (as $$\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4}$$), they are equal. Other options do not satisfy for $$r = 2$$.

Hence, the correct answer is Option A.