Understanding the connection between $\mathbb{P}^2_\mathbb{R}$ and $\mathbb{R}^2$ with example.

Question

I have a problem of understanding the connection between $\mathbb{P}^2_\mathbb{R}$ and $\mathbb{R}^2$.
Let $g:=\{(x,y)\in\mathbb{R}^2\mid x+y-1=0\}$, $l:=\{(x,y)\in\mathbb{R}^2\mid 2x-y-5=0\}$. $\mathbf{g\cap l =(2,-1)}$.
Now let $G:=\{(x:y:z)\in\mathbb{P}^2_\mathbb{R}\mid x+y-z=0\}$, $L:=\{(x:y:z)\in\mathbb{P}^2_\mathbb{R}\mid 2x-y-5z=0\}$ Calculating $G\cap L$: $x+y=z$ $(I)$, $2x-y-5z=0$ $(II)$.$(I)$ in $(II)$: $2x-y-5(x+y)=0 \Leftrightarrow x=-\frac{3}{4}y$. Putting $x=-\frac{3}{4}y$ in $(I)$: $-\frac{3}{4}y+y=z\Leftrightarrow y=4z$. So $G\cap L=\{(-3t:4t:t)\mid t\in\mathbb{R}\setminus\{0\}\}$. So with $t=1$ it follows that $(-3:4:1)\in\mathbb{P}^2_\mathbb{R}$ and $(-3,4)\in g$, but $(-3,4)\notin l$.
I thought that $G,R$ define a set of straight lines through $0$ in $\mathbb{R}^3$. So $G,R$ define planes through $0$ in $\mathbb{R}^3$. The Intersection of these two planes is a straight line which intersects the $\mathbb{R}^2$ ($z=1$) in the same point as $g$ intersects $l$.
Am I wrong with this understanding or where is my mistake?
Thanks in advance, Matthias
Edit:
If I calculate $I-II$ I get $x=2z$ and $y=-z$. So $G\cap L$ would be $\{(2t,-t,t)\mid t\in\mathbb{R}\setminus\{0\}\}$. In this case it follows with $t=1$ that $(2:-1:1)\in\mathbb{P}^2_\mathbb{R} = (2,-1)\in\mathbb{R}^2$. So I think my understanding was occrect, but why do I get different results?