Answer
$(\frac{1}{2\sqrt 3},\frac{3}{2\sqrt 3},\frac{1}{2\sqrt 3},\frac{1}{2\sqrt 3}),(-\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}),(\frac{1}{\sqrt 6},0,\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})$
Work Step by Step
Let $v_1=(1,3,-1,1)\\
v_2=(-1,1,1,-1) \\
v_3=(1,0,2,1)$
Check: $(v_1,v_2)=((1,3,-1,1),(-1,1,1,-1))=1.(-1)+3.1+(-1).1+1.(-1)=0 \\
(v_1,v_3)=((1,3,-1,1),(1,0,2,1))=(-1).1+3.0+(-1).2+1.1=0 \\
(v_2,v_3)=((-1,1,1,-1),(1,0,2,1))=(-1).1+1.0+1.2+1.(-1)=0$
Hence, $v_1,v_2$ and $v_3$ are orthogonal vectors in $R^3$
To determine an orthogonal set, we obtain:
$$\frac{v_1}{||v_1||}=\frac{(1,3,-1,1)}{||(1,3,-1,1)||}=\frac{(1,3,-1,1)}{\sqrt 1^2+3^2+(-1)^2+1^2}=\frac{(1,3,-1,1)}{2\sqrt 3}=(\frac{1}{2\sqrt 3},\frac{3}{2\sqrt 3},\frac{1}{2\sqrt 3},\frac{1}{2\sqrt 3})$$
$$\frac{v_2}{||v_2||}=\frac{(-1,1,1,-1)}{||(-1,1,1,-1)||}=\frac{(-1,1,1,-1)}{\sqrt 1^2+(-1)^2+1^2}=\frac{(-1,1,1,-1)}{\sqrt 3}=(-\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2})$$
$$\frac{v_3}{||v_3||}=\frac{(1,0,2,1)}{||(1,0,2,1)||}=\frac{(1,0,2,1)}{\sqrt 1^2+0^2+2^2+1^2}=\frac{(1,0,2,1)}{\sqrt 6}=(\frac{1}{\sqrt 6},0,\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})$$
Hence, a corresponding orthonormal set of vector is $(\frac{1}{2\sqrt 3},\frac{3}{2\sqrt 3},\frac{1}{2\sqrt 3},\frac{1}{2\sqrt 3}),(-\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}),(\frac{1}{\sqrt 6},0,\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})$
