Let the foci of a hyperbola $$H$$ coincide with the foci of the ellipse $$E : \frac{(x-1)^2}{100} + \frac{(y-1)^2}{75} = 1$$ and the eccentricity of the hyperbola $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\alpha$$ and the length of its conjugate axis is $$\beta$$, then $$3\alpha^2 + 2\beta^2$$ is equal to

Ellipse: $$\frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$. $$a^2=100, b^2=75$$. $$c^2=25$$, $$c=5$$. $$e_E = 1/2$$. Foci: $$(1\pm5, 1) = (6,1)$$ and $$(-4,1)$$.

Hyperbola eccentricity $$e_H = 1/e_E = 2$$. Same foci, centre at (1,1).

For hyperbola: $$c_H = 5$$ (same foci), $$e_H = 2$$, so $$a_H = c_H/e_H = 5/2$$.

$$b_H^2 = c_H^2 - a_H^2 = 25-25/4 = 75/4$$.

Transverse axis $$\alpha = 2a_H = 5$$, conjugate axis $$\beta = 2b_H = 2\sqrt{75/4} = \sqrt{75} = 5\sqrt{3}$$.

$$3\alpha^2 + 2\beta^2 = 3(25) + 2(75) = 75+150 = 225$$.

The correct answer is Option 4: 225.