Answer: 24

Small Hint:

The line already bisects the circle centered at $(17,76)$

Big Hint:

For the other two equal circles, their signed distances from the line must be opposites

Solution:

For a circle of radius $3,$ let $G(d)$ be the signed difference between the areas on the two sides of a line when the center’s signed distance from the line is $d.$ The function $G$ is odd and strictly increasing for $-3<d<3,$ and is constant only after the line no longer cuts the circle. The circle centered at $(17,76)$ contributes zero, so the line must separate the other two centers.

Let $M=(\frac{33}{2},88)$ be the midpoint of those two centers. Among the lines through $(17,76)$ that separate them, the smaller of their two distances to the line is largest when the distances are equal, namely for the line through $M.$ Even then, the distance is $\frac{56}{\sqrt{577}}<3,$ so the line must cut at least one of the two circles. Balance then forces it to cut both. The strict increase of $G$ therefore forces the two signed distances to be opposites, so the required line passes through $M.$ Its slope is $$\frac{88-76}{\frac{33}{2}-17}=-24.$$ Its absolute value is $24.$