Answer: 108

Small Hint:

Segment $AB$ is vertical while its image $A'B'$ is horizontal, so the rotation must be by $90$ degrees

Big Hint:

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a);$ match $A \mapsto A'$

Solution:

The vector $\overrightarrow{AB} = (0, 12)$ is vertical, while $\overrightarrow{A'B'} = (12, 0)$ is horizontal and of the same length, so the rotation turns directions by $90^\circ$ clockwise, and $m = 90.$

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a).$ Applying this to $A = (0, 0)$ and setting the image equal to $A' = (24, 18)$ gives $a - b = 24$ and $a + b = 18,$ so $a = 21$ and $b = -3.$ Checking the other vertices: $B = (0, 12)$ maps to $(21 + 12 + 3,\ -3 + 21)$ $= (36, 18) = B',$ and $C = (16, 0)$ maps to $(21 + 3,\ -3 - 16 + 21)$ $= (24, 2) = C'.$

Therefore $m + x + y = 90$ $+ 21$ $+ (-3) = 108.$