Q:

Projective Plane with three points

I am reading some texts on projective geometry but I am still confused about some easy exercises. I found the following one:
$P_1=[0:1:2:3], P_2=[0:1:2:4], P_3=[1:1:1:1]$ are three points in $\mathbb RP^3$
Now there two things I would like to show that $E=<P_1,P_2,P_3>_{proj}$ is a projective plane and I would like to determine $a_i$ such that $E=\{[x_0:x_1:x_2:x_3]\in\mathbb RP^3:a_0x_0+a_1x_1+a_2x_2+a_3x_3=0\}$
I think the first thing to is to write $<P_1,P_2,P_3>_{proj}$ explicitly as a plane, but I have no idea how to do it, may you could help me with that problem. 

A:

A start: You have $3$ linear equations in the four "unknowns" $a_0,a_1,a_2,a_3$. For example, from the first point we get $0\cdot a_0+1\cdot a_1+2\cdot a_2+3\cdot a_3=0$.  
Find a non-zero solution of the system, in an ad hoc manner, or by row reduction.  As usual in the coordinatized projective world, if $(a_0,a_1,a_2,a_3)$ is a solution, so is $(ka_0,ka_1,ka_2,ka_3)$ for any non-zero $k$.
Added: So we get the system of equations 
$$a_1+2a_2+3a_3=0,\qquad  a_1+2a_2+4a_3=0,\qquad a_0+a_1+a_2+a_3=0.$$
 From the first two equations, by subtraction, $a_3=0$. Let $a_2=t$. Then from the first equation $a_1=-2t$. From the third equation, we have $a_0-2t+t=0$, so $a_0=t$. That gives the general solution $(t,-2t,t,0)$. Choose for example $t=1$ (any non-zero $t$ will do). We conclude that we can take $a_0=1$, $a_1=-2$, $a_2=1$, $a_3=0$. We took advantage of the fact that the numbers we were given are very simple. For messier numbers, formal row-reduction might be better. Certainly it would be if we step up dimensions by $1$ or $2$.     

