Let the hyperbola $$H : \frac{x^2}{a^2} - y^2 = 1$$ and the ellipse $$E : 3x^2 + 4y^2 = 12$$ be such that the length of latus rectum of $$H$$ is equal to the length of latus rectum of $$E$$. If $$e_H$$ and $$e_E$$ are the eccentricities of $$H$$ and $$E$$ respectively, then the value of $$12(e_H^2 + e_E^2)$$ is equal to ______.

The ellipse $$E: 3x^2 + 4y^2 = 12$$ can be written as $$\frac{x^2}{4} + \frac{y^2}{3} = 1$$.

So $$a_E^2 = 4$$, $$b_E^2 = 3$$. The latus rectum of the ellipse is $$\frac{2b_E^2}{a_E} = \frac{2 \times 3}{2} = 3$$.

The eccentricity of the ellipse: $$e_E^2 = 1 - \frac{b_E^2}{a_E^2} = 1 - \frac{3}{4} = \frac{1}{4}$$.

The hyperbola $$H: \frac{x^2}{a^2} - y^2 = 1$$ has $$b_H = 1$$.

The latus rectum of the hyperbola is $$\frac{2b_H^2}{a} = \frac{2}{a}$$.

Setting the latus rectum of $$H$$ equal to that of $$E$$:

$$\frac{2}{a} = 3$$, so $$a = \frac{2}{3}$$.

The eccentricity of the hyperbola: $$e_H^2 = 1 + \frac{b_H^2}{a^2} = 1 + \frac{1}{4/9} = 1 + \frac{9}{4} = \frac{13}{4}$$.

Therefore:

$$12(e_H^2 + e_E^2) = 12\left(\frac{13}{4} + \frac{1}{4}\right) = 12 \times \frac{14}{4} = 12 \times \frac{7}{2} = 42$$

The correct answer is $$42$$.