Let $$S_1 = \{z \in C : |z| \leq 5\}$$, $$S_2 = \left\{z \in C : \text{Im}\left(\frac{z + 1 - \sqrt{3}i}{1 - \sqrt{3}i}\right) \geq 0\right\}$$ and $$S_3 = \{z \in C : \text{Re}(z) \geq 0\}$$. Then the area of the region $$S_1 \cap S_2 \cap S_3$$ is :

$$S_1$$: A disk centered at the origin with radius $$R=5$$. Area $$= 25\pi$$.

$$S_2$$: Let $$w = 1-\sqrt{3}i = 2e^{-i\pi/3}$$. The condition $$\text{Im}(\frac{z+w}{w}) \geq 0$$ simplifies to $$\text{Im}(\frac{z}{w} + 1) \geq 0$$, or $$\text{Im}(z \cdot \bar{w}) \geq 0$$.

 $$z(1+\sqrt{3}i) = (x+iy)(1+\sqrt{3}i) = (x-\sqrt{3}y) + i(y+\sqrt{3}x)$$.

Condition: $$\sqrt{3}x + y \geq 0$$. This is a line through the origin with an angle of $$150^\circ$$ or $$-30^\circ$$.

$$S_3$$: Right half-plane ($$x \geq 0$$).

Intersection: The region is a sector of the circle. $$S_3$$ restricts the angle to $$[-\pi/2, \pi/2]$$. The line from $$S_2$$ has a slope of $$-\sqrt{3}$$ (angle $$-60^\circ$$). The valid region is the sector from $$\theta = -60^\circ$$ to $$\theta = 90^\circ$$.

 Total angle $$= 90^\circ - (-60^\circ) = 150^\circ$$.

 Area $$= \frac{150}{360} \times 25\pi = \frac{5}{12} \times 25\pi = \mathbf{\frac{125\pi}{12}}$$. (Option A)