How many lines can be equidistant from 3 points? How many lines can be drawn in a plane such that they are equidistant from 3 non-collinear points?
@John Bentin has shown below that there are at least 3. Why are there no more than 3?
 A: You can use the following lemma:

If $\ell$ is equidistant from $A$ and $B$ then $\ell$ passes through the midpoint of $AB$ or $\ell$ is parallel to $AB$.
As a consequence, if $\ell$ is equidistant from $A,B$ and $\ell$ separates $A,B$ then $\ell$ passes through the midpoint of $AB$.

A: This had me stumped for a while, because I didn't realize that in geometry, a line is equidistant from a point not if the distance from every point on the line to the point stays the same, but if the line's perpendicular distance from the point stays the same.  Here's an attempt at an explanation:

The three non-colinear points A, B, and C form a triangle.  There is only one point that is equidistant from all three, marked as D...   but there are three lines that are.  To construct one of them:


*

*Draw line BC

*Draw a line from A that intersects BC perpendicularly at point A'.

*Find the midpoint E of the line AA'.


And now the red line parallel to BC and going through E is equidistant to point A and line BC.  You can also think of this line being equidistant to points A and A', or to line BC and its parallel (the red dotted line) passing through point A.
Repeat for lines AB, AC to get the other two lines.
A: Assuming that the line is in the plane of the points, there are three. Consider a triangle ABC. Draw a line parallel to BC so that it is half way between A and BC. The other two are constructed similarly.
(Added for completeness) Consider any line in the plane equidistant from the points.  Suppose for the present that it is not parallel to any side. Then it intersects all three sides of the triangle. The three points of intersection cannot all be internal: say the point D on the line lies on BC produced. But then C would be nearer the line than B, contradicting the given conditions. Hence the supposition can be ruled out. Therefore the line is parallel to a side of the triangle, and so it must be one of the three lines mentioned above.
A: 
If we want $A$ , $B$ and $C$ to be equidistant $L$ must be parallel to $AB$ and pass through midpoint of $CP$
Making line $BC$ we want $B$ and $C$ and $A$ to be equidistant to some line we make the same process draw line perpendicular  on $BC$ and Line parallel to $BC$ and pass through midpoint of the line perpendicular  on $BC$
so we will have three lines such that they are equidistant from these three points
why not more than three?
because we can't draw more than line that is parallel to $AB$ or $BC$ or $AC$
  and pass through midpoint of the line perpendicular  on the line 
