Let $A:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear operator such that \begin{align*} A\left(\begin{bmatrix} 1\\ 3\\ 6 \end{bmatrix}\right) = \begin{bmatrix} 9\\ 28\\ 23 \end{bmatrix},\quad\quad A\left(\begin{bmatrix} -3\\ -5\\ 6 \end{bmatrix}\right) = \begin{bmatrix} 13\\ 44\\ 39 \end{bmatrix},\quad\quad A\left(\begin{bmatrix} 3\\ 3\\ 4 \end{bmatrix}\right) = \begin{bmatrix} 11\\ 24\\ 17 \end{bmatrix} \end{align*}
Find $A\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix}\right)$ for any $\begin{bmatrix} x\\ y\\ z \end{bmatrix}\in\mathbb{R}^3$.
I have calculated $A\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix}\right) = \begin{bmatrix} 3 & -2 & 2\\ 4 & -4 & 6\\ 2 & -3 & 5 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}$.
Next, Equip $\mathbb{R}^3$ with the dot product. Apply the Gram-Schmidt orthogonalization process to the basis for $\mathbb{R}^3, \left\{\begin{bmatrix} 1\\ 3\\ 6 \end{bmatrix},\begin{bmatrix} -3\\ -5\\ 6 \end{bmatrix},\begin{bmatrix} 3\\ 3\\ 4 \end{bmatrix}\right\}$, to obtain an orthonormal basis $B$ for $\mathbb{R}^3$. Find $[A]_B^B$.
I have found the orthonormal basis $B$ for $\mathbb{R}^3$ (shown below), but I don't know what to do next to find $[A]_B^B$.
How I calculated for the orthonormal basis $B$:
Notice that the second and third vectors are already orthogonal. \begin{align*} (-3, -5, 6)\cdot(3, 3, 4) &= (-3)(3) + (-5)(3) + (6)(4)\\ &= -9 - 15 + 24\\ &= 0 \end{align*} Therefore, we only need a third vector: \begin{align*} (-3, -5, 6)\cdot(x,y,z) &= 0\\ -3x-5y+6z &= 0\\\\ (3, 3, 4)\cdot(x, y, z) &= 0\\ 3x + 3y + 4z &= 0\\\\ -3x-5y+6z + 3x + 3y + 4z &= -2y + 10z \end{align*} Assume $z = t$, then $y = 5t$: \begin{align*} -3x - 25t + 6t &= 0\\ x &= -\dfrac{19}{3}t \end{align*} Let $t = 3$, then $x = 19, y = 15, z = 3$. So we get the third vector \begin{align*} (-19, 15, 3) \end{align*} Thus, performing the Gram-Schmidt orthogonalization process, \begin{align*} B &= \left\{\begin{bmatrix} -\frac{3\sqrt{70}}{70}\\-\frac{\sqrt{70}}{14}\\\frac{3\sqrt{70}}{35} \end{bmatrix}, \begin{bmatrix} \frac{3\sqrt{34}}{34}\\\frac{3\sqrt{34}}{34}\\\frac{2\sqrt{34}}{17} \end{bmatrix}, \begin{bmatrix} -\frac{19\sqrt{595}}{595}\\\frac{3\sqrt{595}}{119}\\\frac{3\sqrt{595}}{595} \end{bmatrix}\right\} \end{align*}