TITLE: Construction of the projective plane over $\mathbb{F}_3$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: The points and lines are not triples in $\mathbb {F}^3_3$, but they are elements of $\mathbb{F}^3_2$. So the coordinates of points and lines are in $\mathbb{F}_2$.
For example, your line $L(0,1,1)$ contains the points $[x_0,x_1,x_2]$ with $x_1+x_2=0$. Thus $x_1=x_2=1$ or $x_1=x_2=0$. So the possibilities are:
$[0,1,1]$ , $[1,1,1]$ , $[1,0,0]$.
Edit:
First the question was about the Fano plane, I provided my answer in that case.