1. Identify Key Points and Circles:
- Let $I$ be the incenter of $\triangle ABC$.
- Given $AB = 20$, $BC = 15$, and $BI = 12$.
- Let $CI$ intersect the circumcircle $\omega_1$ of $\triangle ABC$ at $D \neq C$.
- Let $I_a$ and $I_b$ be the excenters opposite $A$ and $B$ respectively.
- Let $E$ be the midpoint of arc $\widehat{BCA}$ on $\omega_1$.
- Let $\rho$ be the circumcircle of $\triangle AIB$.

2. **Properties of Point $D$:**
- It is well-known that $D$ is the center of $\rho$.
- Therefore, $DA = DB = DI = r$, where $r$ is the radius of $\rho$.

3. **Power of Point $X$:**
- Alice draws a line $l$ through $D$ that intersects $\omega_1$ on the minor arc $AC$ at $X$ and the circumcircle $\omega_2$ of $\triangle AIC$ at $Y$ outside $\omega_1$.
- Since $XD^2 - r^2 = XY^2$, $X$ has equal power with respect to $\rho$ and $\omega_2$.

4. Radical Axis and Right Triangle:
- Since $\overline{DE}$ is a diameter of $\omega_1$, $EX \perp DY$.
- Hence, $EX$ is the radical axis of $\rho$ and $\omega_2$.
- It follows that $EY^2 = ED^2 - r^2 = ED^2 - DA^2 = EA^2$ by the Pythagorean Theorem in $\triangle ADE$.

5. **Circle of Diameter $\overline{I_aI_b}$:**
- It is well-known that $A, B, I_a, I_b$ lie on the circle of diameter $\overline{I_aI_b}$ centered at $E$.
- Therefore, $EA = EI_b \implies Y \equiv I_b$.

6. Similarity and Length Calculation:
- Note that $CI_b$ is the external bisector of $\angle BCA \implies \angle BCI_b = 90^\circ + \frac{C}{2} = \angle BIA$.
- Together with $\angle I_bBC = \angle ABI$, this implies $\triangle I_bCB \sim \triangle AIB$.
- Therefore, $\frac{BI_b}{BC} = \frac{BA}{BI}$, whence it follows that $BI_b = \frac{75}{3} = 25$.
- Hence, $II_b = BI_b - BI = 25 - 12 = 13$.

The final answer is $\boxed{13}$.