Let the ellipse $$E:\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$$ and the hyperbola $$H:\frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1$$ have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the value of 24(e+ L) is:

We start by considering the ellipse $$\frac{x^2}{144} + \frac{y^2}{169} = 1$$. Here $$a^2 = 169$$ (since $$b^2 = 169 > 144 = a^2$$, the major axis is along the y-axis), and $$b^2 = 144$$.

Next, consider the hyperbola $$\frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$$, which can be written as $$\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$$, indicating a conjugate hyperbola with transverse axis along the y-axis.

We find the foci of the ellipse: for the ellipse with major axis along y, $$c_E^2 = a^2 - b^2 = 169 - 144 = 25$$, giving $$c_E = 5$$ and foci at $$(0, \pm 5)$$.

For the hyperbola $$\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$$, the focal distance satisfies $$c_H^2 = \lambda^2 + 16$$. Setting $$c_H = 5$$ yields $$25 = \lambda^2 + 16 \implies \lambda^2 = 9$$.

Its eccentricity is $$e = \frac{c_H}{\lambda} = \frac{5}{3}$$.

For the hyperbola $$\frac{y^2}{9} - \frac{x^2}{16} = 1$$, where $$a = 3$$ (transverse semi-axis) and $$b = 4$$, the length of the latus rectum is $$L = \frac{2b^2}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$$.

Finally, computing $$24(e + L)$$ gives $$24\left(\frac{5}{3} + \frac{32}{3}\right) = 24 \times \frac{37}{3} = 8 \times 37 = 296$$.

The answer is Option C: $$296$$.