I don't understand the following proof:
In $P_n(k)$, $n>1$ the Fano plane is satisfied if and only if the characteristic of $k=2$. Using homogeneous coordinates:
Take $A=\{0,0,1\}$, $B=\{1,0,0\}$, $C=\{0,1,0\}$, and $D=\{1,1,1\}$
From here, he can determine the equations of the lines AC, AB and AD. Therefore he has that $P=\{0,1,1\}$, $Q=\{1,1,0\}$, $R=\{1,0,1\}$. How does one obtain these coordinates?
Finally, $P,Q,R$ are on the same line of equation $ax+by+cz=0$. This is equivalent to $a+b=b+c=a+c=0$ which has a no trivial solution if and only if $k$ has characteristic $2$.
I know the basics of the homogeneous coordinate system, and I still can figure out this proof. I would really appreciate if somone explained this proof step by step.
How to obtain $P=\{0,1,1\}$? Well, the line $AC$ is characterized by $x=0$ or equivalently homogeneous line coordinates $\{1,0,0\}$. Likewise $BD$ is $y=z$ or equivalently $\{0,1,-1\}$. You could compute these line coordinates using cross products, or by looking sharply at the numbers, as there is only one homogeneous linear equation (i.e. without constant term) satisfied by both points on each line. So $P$ needs to have $x=0$ and $y=z$, which means $\{0,1,1\}$ and its multiples are the only solution. But multiples of homogeneous coordinates are the same point.
In a similar fashion, $Q$ is characterized by $z=0$ from $BC$ and by $x=y$ from $AD$ leading to $\{1,1,0\}$ and $R$ is characterized by $y=0$ from $AB$ and by $x=z$ from $CD$ leading to $\{1,0,1\}$.
The line $PQ$ could be described by $x+z=y$ or $\{1,-1,1\}$. Likewise $PR$ is $x+y=z$ or $\{1,1,-1\}$. If these two are to be the same line, you need a field where $1=-1$ i.e. characteristic $2$.
Hope this perspective helps you. There are many ways to interpret projective concepts, and this is a guess at what might be useful. Others may come up with completely different explanations, suitable for different visitors of this question.