A plane as a subspace

Subspace. $(0,0,0)\in F$. If $(x,y,z),(x',y',z')\in F$ and $\lambda\in\mathbb{R}$, then
$$(x+\lambda x')+(y+\lambda y')+(z+\lambda z')=(x+y+z)+\lambda(x'+y'+z')=0,$$
so $(x,y,z)+\lambda(x',y',z')\in F$.

Basis. A vector of $F$ has $z=-x-y$, so
$$(x,y,z)=x\,(1,0,-1)+y\,(0,1,-1).$$
The vectors $(1,0,-1)$ and $(0,1,-1)$ span $F$ and are linearly independent (neither is a multiple of the other). They form a basis, and $\dim F=2$.