Orthogonal complement

Equations. $v=(x,y,z)\in F^\perp$ iff:
$$v\cdot(1,1,0)=x+y=0,\qquad v\cdot(0,1,1)=y+z=0.$$
So $y=-x$ and $z=x$, giving $v=x(1,-1,1)$.

Conclusion. $F^\perp=\operatorname{span}((1,-1,1))$, dimension $1$.

Direct sum. $\dim F=2$, $\dim F^\perp=1$, and $2+1=3=\dim\mathbb{R}^3$. Also $F\cap F^\perp=\{0\}$ (a vector in both would be orthogonal to itself, hence zero). So $\mathbb{R}^3=F\oplus F^\perp$.