Let the coefficients of $$x^{-1}$$ and $$x^{-3}$$ in the expansion of $$\left(2x^{1/5} - \frac{1}{x^{1/5}}\right)^{15}$$, $$x > 0$$, be $$m$$ and $$n$$ respectively. If $$r$$ is a positive integer such $$mn^2 = {}^{15}C_r \cdot 2^r$$, then the value of $$r$$ is equal to ______.

We need to find coefficients of $$x^{-1}$$ and $$x^{-3}$$ in $$\left(2x^{1/5} - \frac{1}{x^{1/5}}\right)^{15}$$.

$$T_{r+1} = \binom{15}{r}(2x^{1/5})^{15-r}\left(-\frac{1}{x^{1/5}}\right)^r = \binom{15}{r}2^{15-r}(-1)^r x^{(15-r)/5} \cdot x^{-r/5} = \binom{15}{r}2^{15-r}(-1)^r x^{(15-2r)/5}$$

Setting $$\frac{15-2r}{5} = -1$$ gives $$15 - 2r = -5 \Rightarrow r = 10$$, and hence $$m = \binom{15}{10}2^5(-1)^{10} = \binom{15}{5} \cdot 32 = 3003 \times 32 = 96096$$.

Similarly, $$\frac{15-2r}{5} = -3$$ leads to $$15 - 2r = -15 \Rightarrow r = 15$$, and $$n = \binom{15}{15}2^0(-1)^{15} = -1$$.

Therefore, $$mn^2 = 96096 \times 1 = 96096$$, and $$96096 = \binom{15}{5} \times 32 = 3003 \times 32$$.

We then solve $$\binom{15}{r} \cdot 2^r = 3003 \times 32 = \binom{15}{5} \times 2^5$$, giving $$r = 5$$.

Therefore, the value of $$r = \textbf{5}$$.