Gram-Schmidt orthogonalization

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

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

Step 3. Projections of $v_3$:
$$\langle v_3, u_1 \rangle = 1/\sqrt2, \qquad \langle v_3, u_2 \rangle = (-1+1+2)/\sqrt6 = 2/\sqrt6.$$
So
$$w_3 = v_3 - (1/\sqrt2)u_1 - (2/\sqrt6)u_2 = (0,1,1) - (1/2)(1,1,0) - (1/3)(1,-1,2) = (1/6, 5/6, 1/3).$$
Norm: $\sqrt{1/36+25/36+4/36} = \sqrt{30}/6$, so $u_3 = \sqrt{6/5}(1/6, 5/6, 1/3) = (1/\sqrt{30}, 5/\sqrt{30}, 2/\sqrt{30})$.

Conclusion. The orthonormal basis is
$$u_1 = \frac{1}{\sqrt2}(1,1,0), \quad u_2 = \frac{1}{\sqrt6}(1,-1,2), \quad u_3 = \frac{1}{\sqrt{30}}(1,5,2).$$