Answer
See below
Work Step by Step
Let $x_1=(1,-5,-3) $
According to Gram-Schmidt process, we have:
$v_1=x_1=(2,1,-2)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(0,-1,3)-\frac{)}{||(1,5,-3)||^2}(1,5,-3)\\
=(0,-1,3)-\frac{0.1+(-1).(-5)+3.(-3)}{1^2+(-5)^2+(-3)^2}(1,5,-3)\\
=(0,-1,3)-\frac{4}{35}(1,5,-3)\\
=(0,-1,3)-(\frac{4}{35},-\frac{4}{7},-\frac{12}{35})\\
=(\frac{4}{35},-\frac{11}{7},\frac{93}{35})$
$v_3=x_3-\frac{}{||v_1||^2}v_1-\frac{}{||v_2||^2}v_2-(-6,0,-2)\\
=(-6,0,-2)-\frac{)}{||(1,5,-3)||^2}(1,5,-3)-\frac{)}{||(\frac{4}{35},-\frac{11}{7},\frac{93}{35})||^2}(\frac{4}{35},-\frac{11}{7},\frac{93}{35})\\
=(-6,0,-2)-\frac{-6.1+0.(-5)+(-2).(-3)}{1^2+(-5)^2+(-3)^2}(1,5,-3)-\frac{-6.\frac{4}{35}+0.(-\frac{11}{7})+(-2).\frac{93}{35}}{||(\frac{4}{35})^2+(-\frac{11}{7})^2+(\frac{93}{35})^2}(\frac{4}{35},-\frac{11}{7},\frac{93}{35})\\
=(-6,0,-2)+\frac{105}{167}(\frac{4}{35},-\frac{11}{7},\frac{93}{35})\\
=(\frac{-990}{67},-\frac{165}{167},-\frac{55}{167})$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,5,-3)}{\sqrt 1^2+5^2+(-3)^2}=\frac{(1,5,-3)}{\sqrt 35}=(\frac{1}{35},-\frac{5}{\sqrt 35},-\frac{3}{\sqrt 35})$
$\frac{v_2}{||v_2||}=\frac{(\frac{4}{35},-\frac{11}{7},\frac{93}{35})}{\sqrt \frac{334}{35}}=\frac{35}{334}(\frac{4}{35},-\frac{11}{7},\frac{93}{35})$
$\frac{v_3}{||v_3||}=\frac{(-\frac{990}{167},-\frac{165}{167},-\frac{55}{167})}{\sqrt \frac{6050}{167}}=\sqrt \frac{167}{6050}(-\frac{990}{167},-\frac{165}{167},-\frac{55}{167})$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{1}{\sqrt 35},-\frac{5}{\sqrt 35},-\frac{3}{\sqrt 35}),\sqrt \frac{35}{334}(\frac{4}{35},-\frac{11}{7},\frac{93}{35}),\sqrt \frac{167}{6050}(-\frac{990}{167},\frac{-165}{167},-\frac{55}{167})\}$
