Answer
$B'=\{(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) ,(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}),(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})\}$ is orthonormal basis for $R^3$.
Work Step by Step
Let $B=\{ v_1=(0,1,1),v_2=(1,1,0), v_3=(1,0,1) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(0,1,1)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(1,1,0)- \frac{1}{2}(0,1,1)=(1,\frac{1}{2},-\frac{1}{2})$$
$$w_3=v_3- \frac{v_3\cdot w_1}{w_1\cdot w_1}w_1- \frac{v_3\cdot w_2}{w_2\cdot w_2}w_2\\
=(1,0,1) - \frac{1}{2}(0,1,1)-\frac{1}{3}(1,\frac{1}{2},-\frac{1}{2})=(\frac{2}{3},-\frac{2}{3},\frac{2}{3}).$$
Normalizing $w_1$, $w_2$ and $w_3$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=\frac{1}{\sqrt{2}}(0,1,1)=(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) $$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{\sqrt 3}{\sqrt 2}(\frac{1}{2},-\frac{1}{2})=(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$$
$$u_3=\frac{w_3}{\|w_3\|}=\frac{\sqrt{3}}{2}(\frac{2}{3},-\frac{2}{3},\frac{2}{3})=(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}).$$
Hence, $B'=\{(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) ,(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}),(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})\}$ is orthonormal basis for $R^3$.