Answer
$\{(\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3}),(\frac{4}{\sqrt 42},\frac{5}{\sqrt 42},-\frac{1}{\sqrt 42})\}$
Work Step by Step
Let $x_1=(1,-1,-1) \\
x_2=(2,1,-1)$
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,-1,-1)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(2,1,-1)-\frac{((2,1,-1),(1,2,3))}{||(1,-1,-1)||^2}(1,-1,-1) \\
=(2,1,-1)-\frac{2.1+1.2+(-1).3}{1^2+2^2+3^2}(1,-1,-1)\\
=(\frac{4}{3},\frac{5}{3},-\frac{1}{3})$
To determine an orthogonal set, we obtain:
$\frac{v_1}{||v_1||}=\frac{(1,-1,-1)}{\sqrt 1^2+(-1)^2+(-1)^2}=\frac{(1,-1,-1)}{\sqrt 3}=(\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3})$
$\frac{v_2}{||v_2||}=\frac{(\frac{4}{3},\frac{5}{3},-\frac{1}{3})}{\sqrt (\frac{4}{3})^2+(\frac{5}{3})^2+(-\frac{1}{3})^2}=\frac{(\frac{4}{3},\frac{5}{3},-\frac{1}{3})}{\sqrt \frac{42}{9}}=(\frac{4}{\sqrt 42},\frac{5}{\sqrt 42},-\frac{1}{\sqrt 42})$
Hence, a corresponding orthonormal set of vector is :
$\{(\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3}),(\frac{4}{\sqrt 42},\frac{5}{\sqrt 42},-\frac{1}{\sqrt 42})\}$
