Interpret each set geometrically

We have $S_1=\{z\in\mathbb C:|z|\le 5\}$ which is the closed disc of radius $5$ centered at the origin.

Also, $S_3=\{z\in\mathbb C:\operatorname{Re}(z)\ge 0\}$ is the right half-plane.

The main work is to simplify $S_2$.

Simplify the condition for $S_2$

Let $z=x+iy.$ Then $S_2=\left\{z:\operatorname{Im}\left(\frac{z+1-\sqrt3 i}{1-\sqrt3 i}\right)\ge 0\right\}.$

First write $z+1-\sqrt3 i=(x+1)+i(y-\sqrt3).$

So we need the imaginary part of $\frac{(x+1)+i(y-\sqrt3)}{1-\sqrt3 i}.$

Multiply numerator and denominator by the conjugate $1+\sqrt3 i$:

Find the intersection $S_2\cap S_3$ in angular form

From $S_3$, we have $x\ge 0,$ so we are in the right half-plane, corresponding to angles $-\frac\pi2\le \theta\le \frac\pi2.$

From $S_2$, the region is above the line $y=-\sqrt3 x$. For $x\ge 0$, this means $\tan\theta\ge -\sqrt3,$ so $\theta\ge -\frac\pi3.$

Therefore, inside the disc, the allowed angular region is $-\frac\pi3\le \theta\le \frac\pi2.$

The angular width is $\frac\pi2-\left(-\frac\pi3\right)=\frac{5\pi}{6}.$

Compute the area inside the circle

Since $S_1$ is the disc $r\le 5$, the required region is a sector of radius $5$ and angle $\frac{5\pi}{6}$.

Area of a sector is $\frac12 r^2\theta.$

Hence area

Match with options

$\boxed{\frac{125\pi}{12}}$

This is Option C.