Answer
See below
Work Step by Step
Let $x_1=(2,0,1) $
According to Gram-Schmidt process, we have:
$v_1=x_1=(2,1,-2)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(-3,1,1)-\frac{)}{||(0,2,1)||^2}(0,2,1)\\
=(-3,1,1)-\frac{-3.2+1.0+1.1}{2^2+0^2+1^2}(0,2,1)\\
=(-3,1,1)-\frac{-6+0+1}{4+1}(0,2,1)\\
=(-3,1,1)+(2,0,1)\\
=(-1,1,2)$
$v_3=x_3-\frac{}{||v_1||^2}v_1-\frac{}{||v_2||^2}v_2-(-6,0,-2)\\
=(1,3,-8)-\frac{)}{||(2,0,1)||^2}(2,0,1)-\frac{)}{||(-1,1,2)||^2}(-1,1,2)\\
=(1,3,-8)-\frac{1.2+(-3).0+8.1}{2^2+0^2+1^2}(2,0,1)-\frac{1.(-1)+(-3).1+8.2}{||(-1)^2+1^2+2^2}(-1,1,2)\\
=(1,3,-8)+-2.(2,0,1)-2.(-1,1,2)\\
=(-1,-5,2)$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(2,0,1)}{\sqrt 2^2+0^2+1^2}=\frac{(2,0,1)}{\sqrt 5}=(\frac{2}{\sqrt 5},0,\frac{1}{\sqrt 5})$
$\frac{v_2}{||v_2||}=\frac{(-1,1,2)}{\sqrt 6}=(-\frac{1}{6},\frac{1}{6},\frac{2}{6})$
$\frac{v_3}{||v_3||}=\frac{(-1,-5,2)}{\sqrt 30}=(-\frac{1}{\sqrt 30},-\frac{5}{\sqrt 30},\frac{2}{\sqrt 30})$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{1}{\sqrt 35},-\frac{5}{\sqrt 35},-\frac{3}{\sqrt 35}),\sqrt \frac{35}{334}(\frac{4}{35},-\frac{11}{7},\frac{93}{35}),\sqrt \frac{167}{6050}(-\frac{990}{167},\frac{-165}{167},-\frac{55}{167})\}$
