If the sum of the co-efficients of all the positive even powers of $$x$$ in the binomial expansion of $$\left(2x^3 + \frac{3}{x}\right)^{10}$$ is $$5^{10} - \beta \cdot 3^9$$, then $$\beta$$ is equal to ______.

We need the sum of coefficients of all positive even powers of $$x$$ in the expansion of $$\left(2x^3 + \frac{3}{x}\right)^{10}$$.

First, the general term of the expansion can be written as $$T_{r+1} = \binom{10}{r}(2x^3)^{10-r}\left(\frac{3}{x}\right)^r = \binom{10}{r} \cdot 2^{10-r} \cdot 3^r \cdot x^{30-4r}$$.

Since the power of $$x$$ is $$30 - 4r$$, which is always even, for it to be positive we require $$30 - 4r > 0 \implies r < 7.5 \implies r \leq 7$$. Thus for $$r = 0, 1, 2, \ldots, 7$$ the powers of $$x$$ are $$30, 26, 22, 18, 14, 10, 6, 2$$, all positive and even.

Hence the sum of the required coefficients is $$S = \sum_{r=0}^{7} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r$$.

On the other hand, the total sum of all coefficients, found by setting $$x = 1$$, is $$\sum_{r=0}^{10} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r = (2 + 3)^{10} = 5^{10}$$, so that $$S = 5^{10} - \sum_{r=8}^{10} \binom{10}{r} \cdot 2^{10-r} \cdot 3^r$$.

Computing the excluded terms gives:
$$r = 8: \binom{10}{8} \cdot 2^2 \cdot 3^8 = 45 \cdot 4 \cdot 6561 = 1180980$$
$$r = 9: \binom{10}{9} \cdot 2^1 \cdot 3^9 = 10 \cdot 2 \cdot 19683 = 393660$$
$$r = 10: \binom{10}{10} \cdot 2^0 \cdot 3^{10} = 59049$$
$$\text{Sum} = 1180980 + 393660 + 59049 = 1633689$$

Therefore
$$S = 5^{10} - 1633689 = 5^{10} - \beta \cdot 3^9$$
$$\beta = \frac{1633689}{3^9} = \frac{1633689}{19683} = 83$$
The answer is $$\boxed{83}$$.