Proving line configuration has a single line or an ordinary point for a point configuration in 3-D using Sylvester-Gallai The problem is to prove that if $P$ is a point configuration and $L$ is line configuration in 3-D space that either L contains a single line or there exists an ordinary line/a line that has exactly 2 points from P. 
Now I do know that this is true in a 2-Dimension plane by the theorem by Sylvester-Gallai. However, I am having a hard time visualizing it to 3-D. In an intuitive way, I guess we would be slicing the 3-D region that contains P into 2-D planes and proving it that way with the use of the theorem by Sylvester, but I am not sure exactly how to formulate it. 
Do I just say that let there be some plane that contains more than three points from P and then apply the theorem? In other words, what is the main concepts that an exercise like this would wish to highlight? 
 A: Well, to begin with, you need to be more careful about stating both Sylvester–Gallai and the 3D generalization of it. The correct statement assumes that, while $P$ is an arbitrary configuration of points, $L$ should contain all lines that pass through two or more points from $P$.
Once we've fixed that, there are two approaches to generalizing Sylvester–Gallai to 3D.
First, we can take the proof of the Sylvester–Gallai theorem, and notice that it works in three dimensions just fine. At least, this is true of Kelly's proof: once we choose the point-line pair closest together, every other point or line we draw is going to come from the plane that contains them, so it doesn't matter how many dimensions we have.
Second, we can do what you're trying to do: pick some plane, let $P'$ be the subset of $P$ contained in that plane, and apply Sylvester–Gallai to $P'$. The tricky part is choosing the right plane, because so far our outcomes are:


*

*Sylvester–Gallai tells us that there's a line containing only two points of $P'$. Then we're good, because it also contains only two points of $P$: any point of $P$ on that line also lies on the chosen plane, so it's a point of $P'$.

*Sylvester–Gallai tells us that all points in $P'$ are collinear. This does not necessarily mean that all points in $P$ are collinear. So here is where our choice of plane matters.


Fortunately, if it's the case that not all points of $P$ are collinear, we can pick three noncollinear points in $P$, and take the plane through those. This rules out the second case, and then the proof goes through.
