Construction of the projective plane over $\mathbb{F}_3$ I have a question about constructing  projective plane over $\mathbb{F}_3$. We first establish seven equivalence classes $P= \{ [0,0,1], [0,1,0], [1,0,0], [0,1,1], [1,1,0], [1,0,1], [1,1,1] \}$.
Given a triple $(a_0, a_1, a_2) \in \mathbb{F}^3_3 \setminus (0, 0, 0)$ we define the line
$L(a_0, a_1, a_2)$ as follows: $L(a_0, a_1, a_2) := \{ [x_0; x_1; x_2] \in P : a_0x_0 + a_1x_1 + a_2x_2 = 0 \}$.
It is quite easy for $L(0,0,1), L(0,1,0), L(1,0,0)$, because here we take the points which have zero on the first, second and third coordinate.
It gets more difficult for me for $L(0,1,1)$, because here we need to have $x_1+x_2=0$. Do we treat the coordinates of the points as elements of $\mathbb{F}_2$ or $\mathbb{F}_3$?
There are $26$ nonzero triples in $\mathbb{F}^3_3 $. 
Do we check all 26 $L(x_0, x_1, x_2) $ sets?
Please help, because I really want to understand it.
Thank you.

You are describing a method for getting the Fano plane, which by definition is the projective plane over $\mathbb F_2$.
For getting the projective plane over $\mathbb F_3$ in a similar way, do the following:
Select a set of projective representatives of the nonzero elements of $\mathbb F_3^3$. One way to do this is to select only the vectors whose first nonzero entry is a $1$.
Since $\mathbb F_3$ has $2$ units, you end up with $(3^3 - 1) / 2 = 13$ vectors. Those vectors are called coordinate vectors and give you the $13$ points of the projective plane.
You can also use the coordinate vectors for the description of the lines: Each coordinate vector $v$ corresponds to the line containing all the points $w$ such that the scalar product of $v$ and $w$ equals $0$ (so $\langle v,w\rangle = 0$).
In this way, there are $13$ lines containing $4$ points each.

Typically, the projective plane over a field $K$ is defined in this way:
The subspaces of $K^3$ of dimension $1$ are the points, and the subspaces of $K^3$ of dimension $2$ are the lines.
It may be worth to convince yourself that the above construction is compatible with this description.
