6 . Three circles, each with a radius of 3. The centers are at $(14,92)$, $(17,76)$, and $(19,84)$. Draw a line through the point $(17,76)$ such that the sum of the areas of the parts of the three circles on one side of the line equals the sum of the areas of the parts of the three circles on the other side of the line. Find the absolute value of the slope of this line.
First, note that these three circles are mutually externally tangent equal circles. As shown in the figure, the centers are $\mathrm{O}_{1}(14,92), \mathrm{O}_{2}$ $(19,84), \mathrm{O}_{3}(17, 76)$. Since any line passing through $\mathrm{O}_{3}$ always divides $\odot \mathrm{O}_{3}$ into two equal areas, we only need to consider how to draw a line through a certain point on the plane so that for $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$, the areas on both sides of the line are equal. Clearly, the midpoint $\mathrm{M}(16.5,88)$ of $\mathrm{O}_{1} \mathrm{O}_{2}$ is the center of symmetry for the two equal circles $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$. Any line drawn through this point will meet the above requirement. Therefore, the line passing through $\mathrm{M}(16.5,88)$ and $\mathrm{O}_{3}(17,76)$ is the one we are looking for. Its slope $\mathrm{k}=\frac{88-76}{16.5-17}=-24$. From the figure, we can see that the required line is unique, and the absolute value of its slope is 24.