Completing a family into a basis

Span of the given family. $\operatorname{span}((1,1,0),(0,1,1))$ is the plane $x-y+z=0$ (check: a vector $(x,y,z)$ in the span satisfies $x-y+z=0$).

Choose a vector outside. Take $e_3=(0,0,1)$: it does not satisfy $0-0+1=1\neq0$, so it is outside the span.

Independence. Suppose $\lambda(1,1,0)+\mu(0,1,1)+\nu(0,0,1)=0$. The first coordinate gives $\lambda=0$; the second gives $\lambda+\mu=0$, so $\mu=0$; the third gives $\mu+\nu=0$, so $\nu=0$. Independent.

Conclusion. $((1,1,0),(0,1,1),(0,0,1))$ is a basis of $\mathbb{R}^3$.