If the eccentricity $$e$$ of the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, passing through $$(6, 4\sqrt{3})$$, satisfies $$15(e^2 + 1) = 34e$$, then the length of the latus rectum of the hyperbola $$\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$$ is:

$$15(e^2 + 1) = 34e \implies 15e^2 - 34e + 15 = 0$$

$$\implies (5e - 3)(3e - 5) = 0$$

$$e = \frac{5}{3}$$ ($$e > 1$$)

$$b^2 = a^2\left(\frac{25}{9} - 1\right) \implies b^2 = \frac{16a^2}{9} \quad \text{--- (1)}$$

The hyperbola passes through the point $$(6, 4\sqrt{3})$$:

$$\frac{6^2}{a^2} - \frac{(4\sqrt{3})^2}{b^2} = 1 \implies \frac{36}{a^2} - \frac{48}{b^2} = 1$$

$$\frac{36}{a^2} - \frac{48}{\left(\frac{16a^2}{9}\right)} = 1$$

$$\frac{36}{a^2} - \frac{48 \times 9}{16a^2} = 1 \implies \frac{36}{a^2} - \frac{27}{a^2} = 1 \implies \frac{9}{a^2} = 1$$

$$\therefore a^2 = 9$$

$$b^2 = \frac{16(9)}{9} = 16$$

$$\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$$

Comparing this with the standard form $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$:

$$A^2 = b^2 = 16$$

$$B^2 = 2(a^2 + 1) = 2(9 + 1) = 20$$

The formula for the length of the latus rectum is: $$\text{Length} = \frac{2B^2}{A} = \frac{2(20)}{\sqrt{16}} = \frac{40}{4} = 10$$