Answer
$S=\{(\frac{1}{\sqrt 15},\frac{2}{\sqrt 15},\frac{-1}{\sqrt 15},0,\frac{3}{\sqrt 15}),(\frac{1}{\sqrt 7},\frac{1}{\sqrt 7},0,\frac{2}{\sqrt 7},-\frac{1}{\sqrt 7}),(\frac{4}{\sqrt 77},\frac{2}{\sqrt 77},\frac{-4}{\sqrt 77},\frac{-5}{\sqrt 77},\frac{-4}{\sqrt 77}) \}$
Work Step by Step
Let $v_1=(1,2,-1,0,3)\\
v_2=(1,1,0,2,-1) \\
v_3=(4,2,-4,-5,-4)$
Check: $(v_1,v_2)=((1,2,-1,0,3),(1,1,0,2,-1) )=1.1+2.1+(-1).0+0.2+3(-1)=0 \\
(v_1,v_3)=((1,2,-1,0,3),(4,2,-4,-5,-4))=1.4+2.2+(-1).(-4)+(-5).0+3.(-4)=0 \\
(v_2,v_3)=((1,1,0,2,-1),(4,2,-4,-5,-4))=1.4+2.1+(-4).0+(-5).2+(-1).(-4)=0$
Hence, $v_1,v_2$ and $v_3$ are orthogonal vectors in $R^5$
To determine an orthogonal set, we obtain:
$$\frac{v_1}{||v_1||}=\frac{(1,2,-1,0,3)}{||(1,2,-1,0,3)||}=\frac{(1,2,-1,0,3)}{\sqrt 1^2+2^2+3^2+(-1)^2+0}=\frac{(1,2,-1,0,3)}{\sqrt 15}=(\frac{1}{\sqrt 15},\frac{2}{\sqrt 15},\frac{-1}{\sqrt 15},0,\frac{3}{\sqrt 15})$$
$$\frac{v_2}{||v_2||}=\frac{(1,1,0,2,-1)}{||(1,1,0,2,-1)||}=\frac{(1,1,0,2,-1)}{\sqrt 1^2+1^2+0+2^2+(-1)^2}=\frac{(1,1,0,2,-1)}{\sqrt 7}=(\frac{1}{\sqrt 7},\frac{1}{\sqrt 7},0,\frac{2}{\sqrt 7},-\frac{1}{\sqrt 7})$$
$$\frac{v_3}{||v_3||}=\frac{(4,2,-4,-5,-4)}{||(4,2,-4,-5,-4)||}=\frac{(4,2,-4,-5,-4)}{\sqrt 4^2+2^2+(-4)^2+(-5)^2+(-4)^2}=\frac{(4,2,-4,-5,-4)}{\sqrt 77}=(\frac{4}{\sqrt 77},\frac{2}{\sqrt 77},\frac{-4}{\sqrt 77},\frac{-5}{\sqrt 77},\frac{-4}{\sqrt 77})$$
Hence, a corresponding orthonormal set of vector is $S=\{(\frac{1}{\sqrt 15},\frac{2}{\sqrt 15},\frac{-1}{\sqrt 15},0,\frac{3}{\sqrt 15}),(\frac{1}{\sqrt 7},\frac{1}{\sqrt 7},0,\frac{2}{\sqrt 7},-\frac{1}{\sqrt 7}),(\frac{4}{\sqrt 77},\frac{2}{\sqrt 77},\frac{-4}{\sqrt 77},\frac{-5}{\sqrt 77},\frac{-4}{\sqrt 77}) \}$
