Prove there exists a $2$-coloring of the points of the projective plane of order $11$ such that every line has at least two points of each color To clarify, the projective plane of order $11$ has $133$ lines and $133$ points; any two lines intersect in one point, and any two points determine one line; there are exactly $12$ points on each line, and each point lies on exactly $12$ lines.
I've been thinking about the above problem for a while now, and I believe that I need to use pigeonhole principle. However, I'm not sure how to use pigeonhole principle to prove the existence of a special coloring rather than a guaranteed property of any coloring.
I've also looked at the smaller case of the projective plane of order $3$**, and it seems that this plane has no $2$-coloring such that each line has at least $1$ point of each color. If that's true, I'm not sure how to prove it, although I believe that the strategy to prove this case could help me solve the original problem.
I would appreciate any help, especially a hint or approach that I can use to find the rest of the solution on my own!
EDIT*: I sketched a potential solution in the comments below; please feel free to critique it!
EDIT**: I meant order $2,$ with $7$ points and $7$ lines. Sorry about the confusion there!
 A: Reposted from the comments for more visibility:
Let $X$ be the discrete random variable that counts the number of lines that have at least 2 points of each color. Let $X_i$ be the indicator variable for line $i$.
Then $\mathbb{E}(X) = \mathbb{E}(\sum_{i=1}^{133} X_i) = \sum_{i=1}^{133} \mathbb{E}(X_i) = 133(1−\frac{26}{2^{12}}) = 133 − \frac{1729}{2048} > 132.$ There exists a 1-coloring where $X=0$, but $\mathbb{E}(X) > 132$, so there must exist a 2-coloring where $X=133$. Therefore, there exists a 2-coloring such that all 133 lines each have at least 2 points of each color.
A: Choose four lines, no three of which are concurrent. Call these the chosen lines, and call the other 129 lines ordinary lines.
Provisionally, color a point red if it lies on exactly one chosen line, blue otherwise. (Some colors will be changed later.)
Each chosen line has $9$ red points and $3$ blue points.
Each ordinary line has at least $8$ blue points.
Most ordinary lines have at least $2$ red points. The exceptions are lines which pass through two points, each of which is the intersection of two chosen lines. There are $3$ such lines. On each of those $3$ exceptional lines, choose two points which lie on no chosen line, and recolor them red.
In the revised coloring, every line has at least $2$ red points. The chosen lines are unaffected, as the recolored points do not lie on any chosen line. And since each ordinary line had at least $8$ blue points to start with, and at most $6$ points have been recolored, each ordinary line still has at least $2$ blue points.
P.S. This construction actually achieves a slightly better result: every line has at least $2$ red points and at least $3$ blue points. This is because the $6$ recolored points can't all be collinear, so no line loses too many blue points in the recoloring. In fact, we don't have to recolor more than $4$ points: if the $3$ exceptional lines are concurrent, recolor the point of concurrency and one more point on each of the $3$ lines; otherwise just recolor the $3$ points where the exceptional lines intersect in pairs.
