Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, and $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$, where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$.
1. **Finding the center of rotation:**
   - The center of rotation lies on the intersection of the perpendicular bisectors of the segments \(AA'\), \(BB'\), and \(CC'\). We will find the perpendicular bisectors of \(AA'\) and \(BB'\) and solve for their intersection.

2. **Perpendicular bisector of \(AA'\):**
   - Midpoint of \(AA'\):
     \[
     \left(\frac{24+0}{2}, \frac{18+0}{2}\right) = (12, 9)
     \]
   - Slope of \(AA'\):
     \[
     \frac{18-0}{24-0} = \frac{3}{4}
     \]
   - Slope of the perpendicular bisector:
     \[
     -\frac{4}{3}
     \]
   - Equation of the perpendicular bisector:
     \[
     y = -\frac{4x}{3} + b
     \]
   - Plugging in the midpoint \((12, 9)\):
     \[
     9 = -\frac{4 \cdot 12}{3} + b \implies 9 = -16 + b \implies b = 25
     \]
   - Final equation:
     \[
     y = -\frac{4x}{3} + 25
     \]

3. **Perpendicular bisector of \(BB'\):**
   - Midpoint of \(BB'\):
     \[
     \left(\frac{36+0}{2}, \frac{18+12}{2}\right) = (18, 15)
     \]
   - Slope of \(BB'\):
     \[
     \frac{18-12}{36-0} = \frac{1}{6}
     \]
   - Slope of the perpendicular bisector:
     \[
     -6
     \]
   - Equation of the perpendicular bisector:
     \[
     y = -6x + c
     \]
   - Plugging in the midpoint \((18, 15)\):
     \[
     15 = -6 \cdot 18 + c \implies 15 = -108 + c \implies c = 123
     \]
   - Final equation:
     \[
     y = -6x + 123
     \]

4. **Finding the intersection of the two perpendicular bisectors:**
   - Equate the two equations:
     \[
     -6x + 123 = -\frac{4x}{3} + 25
     \]
   - Clear the fraction by multiplying through by 3:
     \[
     -18x + 369 = -4x + 75
     \]
   - Combine like terms:
     \[
     -18x + 4x = 75 - 369 \implies -14x = -294 \implies x = 21
     \]
   - Substitute \(x = 21\) back into one of the equations to find \(y\):
     \[
     y = -6 \cdot 21 + 123 = -126 + 123 = -3
     \]
   - The center of rotation is \((21, -3)\).

5. **Finding the angle of rotation \(m\):**
   - Use the Law of Cosines to find the angle between \(C\) and \(C'\) with respect to the center of rotation \(O\).
   - Calculate distances:
     \[
     CO = \sqrt{(16-21)^2 + (0+3)^2} = \sqrt{(-5)^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}
     \]
     \[
     C'O = \sqrt{(24-21)^2 + (2+3)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}
     \]
     \[
     CC' = \sqrt{(16-24)^2 + (0-2)^2} = \sqrt{(-8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}
     \]
   - Apply the Law of Cosines:
     \[
     CC'^2 = CO^2 + C'O^2 - 2 \cdot CO \cdot C'O \cdot \cos(m)
     \]
     \[
     68 = 34 + 34 - 2 \cdot 34 \cdot \cos(m)
     \]
     \[
     68 = 68 - 68 \cos(m) \implies 0 = -68 \cos(m) \implies \cos(m) = 0
     \]
   - Since \(\cos(m) = 0\), \(m = 90^\circ\).

6. **Summing \(m\), \(x\), and \(y\):**
   \[
   m + x + y = 90 + 21 - 3 = 108
   \]

The final answer is \(\boxed{108}\)