Let the eccentricity of the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be $$\frac{5}{4}$$. If the equation of the normal at the point $$\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$$ on the hyperbola is $$8\sqrt{5}x + \beta y = \lambda$$, then $$\lambda - \beta$$ is equal to ______.

We have the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e = \frac{5}{4}$$. Since $$e^2 = 1 + \frac{b^2}{a^2}$$ and $$e^2 = \frac{25}{16}$$, it follows that $$\frac{25}{16} = 1 + \frac{b^2}{a^2}$$, which gives $$\frac{b^2}{a^2} = \frac{9}{16}$$.

The point $$\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$$ lies on the hyperbola, so substituting into its equation gives:
$$\frac{64}{5a^2} - \frac{144}{25b^2} = 1$$
Since $$b^2 = \frac{9a^2}{16}$$, this becomes:
$$\frac{64}{5a^2} - \frac{144 \times 16}{25 \times 9a^2} = 1$$
$$\frac{64}{5a^2} - \frac{256}{25a^2} = 1$$
$$\frac{320 - 256}{25a^2} = 1 \implies \frac{64}{25a^2} = 1$$
Therefore $$a^2 = \frac{64}{25}$$ and $$b^2 = \frac{9}{16} \times \frac{64}{25} = \frac{36}{25}$$.

The normal to the hyperbola at $$(x_0,y_0)$$ is given by $$\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2$$. At $$(x_0,y_0) = \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$$ this becomes:
$$\frac{(64/25)x}{8/\sqrt{5}} + \frac{(36/25)y}{12/5} = \frac{64}{25} + \frac{36}{25}$$
$$\frac{64\sqrt{5}}{200}x + \frac{36 \times 5}{25 \times 12}y = \frac{100}{25}$$
$$\frac{8\sqrt{5}}{25}x + \frac{3}{5}y = 4$$
Multiplying through by 25 yields $$8\sqrt{5}\,x + 15y = 100$$.

Comparing with the given form $$8\sqrt{5}\,x + \beta y = \lambda$$ shows that $$\beta = 15$$ and $$\lambda = 100$$, hence $$\lambda - \beta = 100 - 15 = 85$$. Therefore, the answer is $$\boxed{85}$$.