Let, $x_{1} = (1,2,0) \\ x_{2} = (0,3,1)$

we can extend it to basis for $R^3$ by adding one vector from the standard basis

Thus $x_{3} = (1,0,0)$

$\begin{vmatrix} 1 & 2 & 0\\ 0 & 3 & 1 \\ 1 & 0 & 0 \end{vmatrix} = 0 - 2(-1) + 0 = 2 \neq 0$

Using the Gram-Schmidt process we orthogonize the basis.

$x_{1} = (1,2,0) \hspace{1cm} x_{2} = (0,3,1) \hspace{1cm} x_{3} = (1,0,0)$

$$u_{1} = \frac{1 [1 \hspace{0.25cm} 2 \hspace{0.25cm} 0]}{\sqrt{1^2 + 2^2 + 0^2}} \\ = \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$$

$v_{1} = x_{1} = (1,2,0) \\ v_{2} = x_{2} - \frac{\prec x_{2}, v_{1} \succ}{\prec v_{1}, v_{1} \succ} v_{1}$

$$\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix} - \Bigg\{\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix} \cdot \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}\Bigg\} \times \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$$

Now first solving the Dot product i.e. curly bracket

$$v_{2} = \begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix} - \frac{1}{5} [0 + 6 + 0] \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \\ v_{2} = \begin{bmatrix} 0 \\3 \\ 1 \end{bmatrix} - \frac{1}{5} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \\ v_{2} = \bigg[\frac{-6}{5}, \frac{3}{5}, 1 \bigg]$$

Now, $v_{2} = \bigg[\frac{-6}{5}, \frac{3}{5}, 1 \bigg]$

$$u_{2} = \frac{\bigg[\frac{-6}{5}, \frac{3}{5}, 1 \bigg]}{\sqrt{\bigg(\frac{-6}{5}\bigg)^2 + \bigg(\frac{3}{5}\bigg)^2 + 1^2}} \\ u_{2} = \frac{5}{\sqrt{70}} \bigg[\frac{-6}{5}, \frac{3}{5}, 1 \bigg]$$

$$\therefore v_{3} = x_{3} - \frac{\prec x_{3}, v_{1} \succ}{\prec v_{1}, v_{1} \succ} v_{1} - \frac{\prec x_{3}, v_{2} \succ}{\prec v_{2}, v_{2} \succ} v_{2} \\ = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \Bigg\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \cdot \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \Bigg\} \cdot \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} - \Bigg\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \cdot \frac{5}{\sqrt{70}} \begin{bmatrix} \frac{-6}{5} \\ \frac{3}{5} \\ 1 \end{bmatrix} \Bigg \} \cdot \frac{5}{\sqrt{70}} \begin{bmatrix} \frac{-6}{5} \\ \frac{3}{5} \\ 1 \end{bmatrix} \\ = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \frac{1}{5} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} - \frac{25}{70} \bigg[ \frac{-6}{5} + 0 + 0 \bigg] \begin{bmatrix} \frac{-6}{5} \\ \frac{3}{5} \\ 1 \end{bmatrix} \\ = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{5} \\ \frac{2}{5} \\ 0 \end{bmatrix} + \frac{30}{70} \begin{bmatrix} \frac{-6}{5} \\ \frac{3}{5} \\ 1 \end{bmatrix} \\ = \begin{bmatrix} \frac{2}{7} \\ \frac{-1}{7} \\ \frac{3}{7} \end{bmatrix} $$

$\therefore v_{1} = [1,2,0] \hspace{1cm} v_{2} = \bigg[\frac{-6}{5}, \frac{3}{5}, 1\bigg]$

$v_{3} = \bigg[\frac{2}{7}, \frac{-1}{7}, \frac{3}{7} \bigg] $is an orthogonal basis.