It is from Chapter 8, Example 12 in Linear Algebra by Hoffman and Kunze.
Consider the vectors $$ \begin{gather*} \beta_1 =& (3,0,4) \\ \beta_2 =& (-1,0,7) \\ \beta_3 =& (2,9,11) \end{gather*} $$ Apply the Gram-Schmidt process to $\beta_1,\beta_2,\beta_3$, we obtain $\alpha_1 = (3,0,4), \alpha_2= (-4,0,3), \alpha_3 = (0,9,0)$, which is an orthogonal basis for $\mathbf{R}^3$.
To express an arbitrary vector $(x_1,x_2,x_3)$ in $\mathbf{R}^3$ as an linear combination of $\alpha_1, \alpha_2, \alpha_3$, it is not necessary to solve any linear equations, we can use the following result:
Corollary. If a vector $ \beta $ is a linear combination of an orthogonal sequence of non-zero vectors $ \alpha _ 1 , \dots , \alpha _ m $, then $ \beta $ is the particular linear combination $$ \beta = \sum _ { k = 1 } ^ m \frac { ( \beta | \alpha _ k ) } { \| \alpha _ k \| ^ 2 } \alpha _ k \text . $$
Then we get an arbitrary vector $(x_1,x_2,x_3)$ as a linear combination of $\alpha_1, \alpha_2, \alpha_3$ as follows, using standard inner product: $$(x_1,x_2,x_3) = \frac{3x_1+4x_3}{25}\alpha_1+\frac{-4x_1+3x_3}{25}\alpha_2+\frac{x_2}{9}\alpha_3$$
However, how to understand the following argument:
To put this point in another way, what we have shown is the following: the basis $ \{ f _ 1 , f _ 2 , f _ 3 \} $ of $ \left ( \mathbf R ^ 3 \right ) ^ * $ which is dual to the basis $ \{ \alpha _ 1 , \alpha _ 2 , \alpha _ 3 \} $ is defined explicitly by the equations $$ \begin{gather*} f _ 1 ( x _ 1 , x _ 2 , x _ 3 ) =& \frac { 3 x _ 1 + 4 x _ 3 } { 25 } \\ f _ 2 ( x _ 1 , x _ 2 , x _ 3 ) =& \frac { - 4 x _ 1 + 3 x _ 3 } { 25 } \\ f _ 3 ( x _ 1 , x _ 2 , x _ 3 ) =& \frac { x _ 2 } 9 \end{gather*} $$
To find a dual basis, all I know is to define $f_i(\alpha_j) = \delta_{ij}$. So here, for example, if I want to calculate $f_1$, we get $$ \begin{gather*} f_1(\alpha_1) =& 1 \\ f_1(\alpha_2) =& 0 \\ f_1(\alpha_3) =& 0 \end{gather*} $$
How do I proceed?
Each $f_i$ is some linear function of the coordinates $x_1, x_2, x_3$. That is, there exists some $m_{i1}, m_{i2}, m_{i3}$ such that $$\begin{array}cf_i(x_1, x_2, x_3) = m_{i1}x_1 + m_{i2}x_2 + m_{i3}x_3 = \begin{bmatrix}m_{i1}&m_{i2}&m_{i3}\end{bmatrix}\\\phantom{|}\\\phantom{|} \end{array}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$ We can write all three functions in combination as $$\begin{bmatrix}f_1\\f_2\\f_3\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{bmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\m_{31}&m_{32}&m_{33}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$
Now plugging in the three points and similarly combining the equations gives us: $$\begin{bmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\m_{31}&m_{32}&m_{33}\end{bmatrix}\begin{bmatrix}3&-4&0\\0&0&9\\4&3&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$
Can you take it from there?