Gram-Schmidt orthonormalization

Step 1. $u_1=v_1/\|v_1\|=(1,1,0)/\sqrt2$.

Step 2. Projection of $v_2$ onto $u_1$: $\langle v_2,u_1\rangle=1/\sqrt2$. Orthogonal part:
$$w_2=v_2-\frac{1}{\sqrt2}u_1=(1,0,1)-(1/2)(1,1,0)=(1/2,-1/2,1).$$
Norm: $\sqrt{1/4+1/4+1}=\sqrt{3/2}$, so $u_2=w_2/\|w_2\|=(1,-1,2)/\sqrt6$.

Step 3. Projections of $v_3$:
$$\langle v_3,u_1\rangle=1/\sqrt2,\qquad \langle v_3,u_2\rangle=2/\sqrt6.$$
Orthogonal part:
$$w_3=v_3-\frac{1}{\sqrt2}u_1-\frac{2}{\sqrt6}u_2.$$
Compute: $\frac{1}{\sqrt2}u_1=(1/2,1/2,0)$ and $\frac{2}{\sqrt6}u_2=(1/3,-1/3,2/3)$. So
$$w_3=(0,1,1)-(1/2,1/2,0)-(1/3,-1/3,2/3)=(-5/6,5/6,1/3).$$
Norm: $\sqrt{25/36+25/36+4/36}=\sqrt{54}/6$. Simplify: $\sqrt{54}=3\sqrt6$, so $\|w_3\|=\sqrt6/2$. Hence $u_3=w_3/\|w_3\|=(-5,5,2)/\sqrt{54}$.

Conclusion. The orthonormal basis is
$$u_1=\frac{1}{\sqrt2}(1,1,0),\quad u_2=\frac{1}{\sqrt6}(1,-1,2),\quad u_3=\frac{1}{\sqrt{54}}(-5,5,2).$$