Answer
See below
Work Step by Step
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,2,0,1)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(2,1,1,0)-\frac{)}{||(1,2,0,1)||^2}(1,2,0,1)\\
=(2,1,1,0)-\frac{2.1+1.2+1.0+0.1}{1^2+2^2+1^2+0^2}(1,2,0,1)\\
=(2,1,1,0)-\frac{2+0+2+0}{2}(1,2,0,1)\\
=(2,1,1,0)-\frac{2}{3}(1,2,0,1)\\
=\frac{1}{3}(4,-1,3,-2)$
$v_3=x_3-\frac{}{||v_1||^2}v_1-\frac{}{||v_2||^2}v_2-(-6,0,-2)\\
=(1,0,2,1)-\frac{)}{||(1,2,0,1)||^2}(1,2,0,1)-\frac{)}{||\frac{1}{3}(4,-1,3,-2)||^2}\frac{1}{3}(4,-1,3,-2)\\
=(1,0,2,1)-\frac{1+0+0+1}{1+4+0+1}(1,2,0,1)-\frac{4+0+6-2}{16+1+9+4}(4,-1,3,-2)\\
=(1,0,2,1)+\frac{1}{3}(1,2,0,1)-\frac{4}{15}(4,-1,3,-2)\\
=\frac{2}{5}(-1,-1,3,3)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,2,0,1)}{\sqrt 1^2+0^2+2^2+1^2}=\frac{(1,2,0,1)}{\sqrt 6}=(\frac{1}{\sqrt 6},\frac{2}{\sqrt 6},0,\frac{1}{\sqrt 6})$
$\frac{v_2}{||v_2||}=\frac{\frac{1}{3}(4,-1,3,-2)}{\sqrt 30}=(\frac{4}{\sqrt 30},-\frac{1}{\sqrt 30},\frac{3}{\sqrt 30},-\frac{2}{\sqrt 30})$
$\frac{v_3}{||v_3||}=\frac{\frac{2}{5}(-1,-1,3,3)}{\sqrt 20}=(-\frac{1}{\sqrt 20},0,-\frac{1}{\sqrt 20},\frac{3}{\sqrt 20},\frac{3}{\sqrt 20}))$
