Orthonormal basis of a subspace

Step 1. $u_1=(1,1,0)/\sqrt2$.

Step 2. Projection of $(1,0,1)$ onto $u_1$: $\langle(1,0,1),u_1\rangle = 1/\sqrt2$. Orthogonal part:
$$(1,0,1) - \frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}(1,1,0) = (1,0,1)-(1/2,1/2,0)=(1/2,-1/2,1).$$
Norm: $\sqrt{1/4+1/4+1}=\sqrt{3/2}$. So
$$u_2 = \sqrt{2/3}(1/2,-1/2,1) = (1/\sqrt6,-1/\sqrt6,\sqrt{2/3}).$$

Conclusion. An orthonormal basis of $F$ is
$$u_1=\frac1{\sqrt2}(1,1,0), \qquad u_2=\frac1{\sqrt6}(1,-1,2).$$