Use Gram-Schmidt process to transform the basis ${U_{1}, U_{2}, U_{3}}$ into orthonormal bases where $U_{1} = \{ 1,1,1 \} $, $U_{2}=\{ 0,1, 1 \}$, $U_{3}=\{ 0, 0, 1 \}$.

Solution:

Step 1: $V_{1}=U_{1}=(1, 1, 1)$

Step 2: $V_{2}=U_{2} - proj U_{2} = U_{2}- \cfrac{\lt U_{2} \cdot V_{1} \gt}{||V_{1}||^{2}} V_{1}$

$\therefore U_{2} \cdot V_{1} = (0,1,1)(1,1,1)=(0+1+1)=2$

$||V_{1}||=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}} = \sqrt{3}$

$||V_{1}||^{2}=3$

$\therefore V_{2} = (0, 1, 1) - \cfrac{(2)}{3} (1,1,1) = (0, 1, 1) - \left( \cfrac{2}{3}, \cfrac{2}{3}, \cfrac{2}{3} \right) =\left[ 0-\cfrac{2}{3}, 1-\cfrac{2}{3}, 1-\cfrac{2}{3} \right]$

$\therefore V_{2} =\left[ -\cfrac{2}{3}, \cfrac{1}{3}, \cfrac{1}{3} \right]$

Step 3: $V_{3}=U_{3}- proj U_{3}=U_{3}- \cfrac{(U_{3} \cdot V_{1})}{||V_{1}||^{2}}V_{1}-- \cfrac{(U_{3} \cdot V_{2})}{||V_{2}||^{2}}V_{2}$

$\therefore (U_{3} \cdot V_{1}) = (0, 0, 1) (1, 1, 1) = 0+0+1 =1$

$||V_{1}|| = \sqrt{1^{2} +1^{2} +1^{2}} = \sqrt{3}$

$||V_{1}||^{2} = 3$

$U_{3} \cdot V_{2} = (0, 0, 1) \left( -\cfrac{2}{3}, \cfrac{1}{3}, \cfrac{1}{3} \right) =0+0+\cfrac{1}{3} = \cfrac{1}{3}$

$||V_{2}||=\sqrt{\left( -\cfrac{2}{3} \right)^{2}+\left( \cfrac{1}{3} \right)^{2}+\left( \cfrac{1}{3} \right)^{2}}=\sqrt{\cfrac{4}{9}+\cfrac{1}{9}+\cfrac{1}{9}}=\sqrt{\cfrac{6}{9}}=\cfrac{\sqrt{6}}{3}$

$||V_{2}||^{2}=\cfrac{6}{9}=\cfrac{2}{3}$

$V_{3}=(0,0,1)-\cfrac{1}{3}(1,1,1)-\cfrac{1/3}{2/3}(-\cfrac{2}{3},\cfrac{1}{3},\cfrac{1}{3})$

$V_{3}=(0,0,1)-(\cfrac{1}{3},\cfrac{1}{3},\cfrac{1}{3})-\cfrac{1}{2}(-\cfrac{2}{3},\cfrac{1}{3},\cfrac{1}{3})$

$V_{3}=(0,0,1)-(\cfrac{1}{3},\cfrac{1}{3},\cfrac{1}{3})-(-\cfrac{2}{6},\cfrac{1}{6},\cfrac{1}{6})$

$V_{3}=(0,-\cfrac{1}{2},\cfrac{1}{2})$

$\therefore V_{1}=(1,1,1), V_{2}=(\cfrac{1}{3},\cfrac{1}{3},\cfrac{1}{3}), V_{3}=(0,-\cfrac{1}{2},\cfrac{1}{2})$

$||V_{1}|| = \sqrt{1^{2} +1^{2} +1^{2}} = \sqrt{3}$

$||V_{2}||=\sqrt{\left( -\cfrac{2}{3} \right)^{2}+\left( \cfrac{1}{3} \right)^{2}+\left( \cfrac{1}{3} \right)^{2}}=\sqrt{\cfrac{4}{9}+\cfrac{1}{9}+\cfrac{1}{9}}=\sqrt{\cfrac{6}{9}}=\cfrac{\sqrt{6}}{3}$

$||V_{3}||=\sqrt{\left( 0 \right)^{2}+\left( -\cfrac{1}{2} \right)^{2}+\left( \cfrac{1}{2} \right)^{2}}=\sqrt{0+\cfrac{1}{4}+\cfrac{1}{4}}=\sqrt{\cfrac{2}{4}}=\cfrac{1}{\sqrt{2}}$

Orthogonal Basis:

$q_{1}=\cfrac{V_{1}}{||V_{1}||}=\cfrac{(1,1,1)}{\sqrt{3}}= \left ( \cfrac{1}{\sqrt{3}}, \cfrac{1}{\sqrt{3}}, \cfrac{1}{\sqrt{3}} \right)$

$q_{2}=\cfrac{V_{2}}{||V_{2}||}=\cfrac{1}{\sqrt{6}/3} \left ( -\cfrac{2}{3}, \cfrac{1}{3}, \cfrac{1}{3} \right) =\cfrac{3}{\sqrt{6}} \left ( -\cfrac{2}{3}, \cfrac{1}{3}, \cfrac{1}{3} \right) = \left ( -\cfrac{2}{\sqrt{6}}, \cfrac{1}{\sqrt{6}}, \cfrac{1}{\sqrt{6}} \right)$

$q_{3}=\cfrac{V_{3}}{||V_{3}||}=\cfrac{1}{1/ \sqrt{2}} \left ( 0, -\cfrac{1}{2}, \cfrac{1}{2} \right)= \sqrt{2} \left ( 0, -\cfrac{1}{2}, \cfrac{1}{2} \right) =\left ( 0, -\cfrac{\sqrt{2}}{2}, \cfrac{\sqrt{2}}{2} \right)=\left ( 0, -\cfrac{1}{\sqrt{2}}, \cfrac{1}{\sqrt{2}}\right) $

Because $(2=\sqrt{2} \cdot \sqrt{2})$