Answer
$(\frac{2}{\sqrt 5},-\frac{1}{\sqrt 5},\frac{1}{\sqrt 5}),(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{-1}{\sqrt 3}),(0,\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$
Work Step by Step
Let $v_1=(2,-1,1)\\
v_2=(1,1,-1) \\
v_3=(0,1,1)$
Check: $(v_1,v_2)=((2,-1,1),(1,1,-1))=2.1+(-1).1+1.(-1)=0 \\
(v_1,v_3)=((2,-1,1),(0,1,1))=2.0+(-1).1+1.1=0 \\
(v_2,v_3)=((1,1,-1),(0,1,1))=1.0+1.1+1.(-1)=0$
Hence, $v_1,v_2$ and $v_3$ are orthogonal vectors in $R^2$
To determine an orthogonal set, we obtain:
$$\frac{v_1}{||v_1||}=\frac{(2,-1,1)}{||(2,-1,1)||}=\frac{(2,-1,1)}{\sqrt 2^2+(-1)^2+1^2}=\frac{(2,-1,1)}{\sqrt 5}=(\frac{2}{\sqrt 5},\frac{-1}{\sqrt 5},\frac{1}{\sqrt 5})$$
$$\frac{v_1}{||v_1||}=\frac{(1,1,-1)}{||(1,1,-1)||}=\frac{(1,1,-1)}{\sqrt 1^2+(-1)^2+1^2}=\frac{(1,1,-1)}{\sqrt 3}=(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{-1}{\sqrt 3})$$
$$\frac{v_3}{||v_3||}=\frac{(0,1,1)}{||(0,1,1)||}=\frac{(0,1,1)}{\sqrt 0^2+1^2+1^2}=\frac{(0,1,1)}{\sqrt 2}=(0,\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$$
Hence, a corresponding orthonormal set of vector is $(\frac{2}{\sqrt 5},-\frac{1}{\sqrt 5},\frac{1}{\sqrt 5}),(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{-1}{\sqrt 3}),(0,\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$
