U = span{(1 1 0), (0 1 1)}
Find a basis for U(prep)
$\displaystyle (x,y,z) \in U^{\perp}$ if and only if $\displaystyle (x,y,z) \cdot (1,1,0)=(x,y,z) \cdot (0,1,1)=0.$ that will give you: $\displaystyle x+y=y+z=0.$ thus: $\displaystyle x=z=-y$ and $\displaystyle y$ can be anything. so $\displaystyle \{(-1,1,-1) \}$ is a beasis for $\displaystyle U^{\perp}. \ \Box$