Giving coordinates in a projective plane When we are giving coordinates to the points of the Fano plane, we do so by giving every point a triplet: $(a_1, a_2,a_3)$ from $\mathbb F_2$ so that if three points are collinear then the pointwise sum of their coordinates $mod \space 2$ is $0$.https://en.wikipedia.org/wiki/Fano_plane#Homogeneous_coordinates
It is the case when we have $2^2+2^1+1=7$ points. But how do we generalise collinearity for projective planes with $n^2+n+1$ points and coordinates from $\mathbb F_n$? (e.g. when are three points collinear when we have $n=4$ so $21$ points and $a_1,a_2,a_3 $ are from $\mathbb F_4$?)
 A: In the general case, remember that each point in the projective plane is represented by $n-1$ different vectors in $\mathbb F_n^3$ (which are all scalar multiples of each other).
$3$ different projective points are collinear exactly if a set containing a representative of each of them span a two-dimensional subspace of $\mathbb F_n^3$.
Another way to express this condition is that if $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ and $(c_1,c_2,c_3)$ represent three (not necessarily different) projective points, then they are collinear iff the matrix
$$ \begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\  a_3 & b_3 & c_3 \end{bmatrix} $$ is singular -- that is, if its determinant is $0$, or in yet other words if $((a_1,a_2,a_3)\times(b_1,b_2,b_3))\cdot(c_1,c_2,c_3)=0$.

Note that this doesn't reduce to the nice simple rule for the Fano plane when $n=2$. The existence of the simple rule depends critically on the fact that every line contains exactly three points, so when you have two different points, you can compute directly what the third one must be: $\vec c = \vec a + \vec b$ is the only element in $\operatorname{span}(\vec a,\vec b)$ that isn't either zero or one of the known points. Since we're in characteristic $2$, this is the same as $\vec a + \vec b + \vec c = 0$.
