We construct three circles: $O$ with diameter $AB$ and area $12+2x$, $P$ with diameter $AC$ and area $24+x$, and $Q$ with diameter $BC$ and area $108-x$.  Given that $C$ is on circle $O$, compute $x$.
1. **Identify the relationship between the circles and their areas:**
   - Circle \( O \) has diameter \( AB \) and area \( 12 + 2x \).
   - Circle \( P \) has diameter \( AC \) and area \( 24 + x \).
   - Circle \( Q \) has diameter \( BC \) and area \( 108 - x \).

2. **Use the fact that \( C \) is on circle \( O \):**
   - Since \( AB \) is the diameter of circle \( O \), angle \( ACB \) is a right angle (by the Inscribed Angle Theorem).

3. **Apply the Pythagorean Theorem:**
   - Since \( \angle ACB \) is a right angle, we have:
     \[
     AC^2 + BC^2 = AB^2
     \]

4. **Relate the areas of the circles to their diameters:**
   - The area of a circle is given by \( \pi r^2 \), where \( r \) is the radius.
   - The radius is half the diameter, so for circle \( O \) with diameter \( AB \):
     \[
     \pi \left(\frac{AB}{2}\right)^2 = 12 + 2x
     \]
   - For circle \( P \) with diameter \( AC \):
     \[
     \pi \left(\frac{AC}{2}\right)^2 = 24 + x
     \]
   - For circle \( Q \) with diameter \( BC \):
     \[
     \pi \left(\frac{BC}{2}\right)^2 = 108 - x
     \]

5. **Express the areas in terms of the diameters:**
   - For circle \( O \):
     \[
     \pi \left(\frac{AB}{2}\right)^2 = 12 + 2x \implies \frac{\pi AB^2}{4} = 12 + 2x \implies \pi AB^2 = 48 + 8x
     \]
   - For circle \( P \):
     \[
     \pi \left(\frac{AC}{2}\right)^2 = 24 + x \implies \frac{\pi AC^2}{4} = 24 + x \implies \pi AC^2 = 96 + 4x
     \]
   - For circle \( Q \):
     \[
     \pi \left(\frac{BC}{2}\right)^2 = 108 - x \implies \frac{\pi BC^2}{4} = 108 - x \implies \pi BC^2 = 432 - 4x
     \]

6. **Substitute the areas into the Pythagorean Theorem:**
   \[
   AC^2 + BC^2 = AB^2
   \]
   - Using the expressions for the areas:
     \[
     \frac{96 + 4x}{\pi} + \frac{432 - 4x}{\pi} = \frac{48 + 8x}{\pi}
     \]
   - Simplify the equation:
     \[
     96 + 4x + 432 - 4x = 48 + 8x
     \]
     \[
     528 = 48 + 8x
     \]
     \[
     480 = 8x
     \]
     \[
     x = 60
     \]

The final answer is \( \boxed{60} \).