Triangle $ABC$ has right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$.  Find $PC$.

[asy]
pair A=(0,5), B=origin, C=(12,0), D=rotate(-60)*C, F=rotate(60)*A, P=intersectionpoint(A--D, C--F);
draw(A--P--B--A--C--B^^C--P);
dot(A^^B^^C^^P);
pair point=P;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$P$", P, NE);[/asy]
1. Given that $\triangle ABC$ is a right triangle with $\angle B = 90^\circ$, and a point $P$ inside the triangle such that $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA = 120^\circ$.
2. Since $\angle APB + \angle BPC + \angle CPA = 360^\circ$ and $\angle APB = \angle BPC = \angle CPA$, we have:
   \[
   \angle APB = \angle BPC = \angle CPA = \frac{360^\circ}{3} = 120^\circ
   \]
3. Applying the Law of Cosines in $\triangle APB$:
   \[
   AB^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos(120^\circ)
   \]
   Since $\cos(120^\circ) = -\frac{1}{2}$, we get:
   \[
   AB^2 = 10^2 + 6^2 - 2 \cdot 10 \cdot 6 \cdot \left(-\frac{1}{2}\right)
   \]
   \[
   AB^2 = 100 + 36 + 60 = 196
   \]
   \[
   AB = \sqrt{196} = 14
   \]
4. Let $PC = x$. Applying the Law of Cosines in $\triangle BPC$:
   \[
   BC^2 = PB^2 + PC^2 - 2 \cdot PB \cdot PC \cdot \cos(120^\circ)
   \]
   \[
   BC^2 = 6^2 + x^2 - 2 \cdot 6 \cdot x \cdot \left(-\frac{1}{2}\right)
   \]
   \[
   BC^2 = 36 + x^2 + 6x
   \]
5. Applying the Law of Cosines in $\triangle CPA$:
   \[
   AC^2 = PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos(120^\circ)
   \]
   \[
   AC^2 = 10^2 + x^2 - 2 \cdot 10 \cdot x \cdot \left(-\frac{1}{2}\right)
   \]
   \[
   AC^2 = 100 + x^2 + 10x
   \]
6. Using the Pythagorean Theorem in $\triangle ABC$:
   \[
   AB^2 + BC^2 = AC^2
   \]
   \[
   14^2 + (36 + x^2 + 6x) = 100 + x^2 + 10x
   \]
   \[
   196 + 36 + x^2 + 6x = 100 + x^2 + 10x
   \]
   \[
   232 + 6x = 100 + 10x
   \]
   \[
   132 = 4x
   \]
   \[
   x = \frac{132}{4} = 33
   \]

The final answer is $\boxed{33}$