Here's how to start thinking about it. I'll use $\mathcal B$ to denote the coordinate representation in the basis $\mathcal B = \left\{(1,1,1),(0,2,2),(0,0,3) \right\}$ and $\mathcal S$ to denote the coordinate representation in the standard basis. The transformation matrix for the basis $\mathcal B$ is given by
$$A = \left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right].$$
This tells you that the first basis vector in $\mathcal B$ is mapped to the third basis vector in $\mathcal B$, the second is mapped to two times the third, and so on. That is,
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}1\\0\\0 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\0\\1 \end{array} \right]_{\mathcal B} = (1)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S},$
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}0\\1\\0 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\0\\2 \end{array} \right]_{\mathcal B} = (2)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S}, \text{ and }$
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}0\\0\\1 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\1\\3 \end{array} \right]_{\mathcal B} = (1)\left[\begin{array}{c}0\\2\\2 \end{array} \right]_{\mathcal S} + (3)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S}.$
So you see that in the standard basis
\begin{equation}
\begin{split}
T \text{ maps } & (1,1,1) \text{ to } (0,0,3), \\
T \text{ maps } & (0,2,2) \text{ to } (0,0,6), \\
T \text{ maps } & (0,0,3) \text{ to } (0,2,11).
\end{split}
\end{equation}
Now see if you can use this information to construct the matrix for $T$ in the standard basis.
You have$$f(b_1)=(4,8,12),\ f(b_2)=(-1,-2,-3)\text{ and }b_3=(3,6,9).$$But, with respect to the basis $B=\{b_1,b_2,b_3\}$, you have$$f(b_1)=\left(\frac{20}3,-\frac{16}3,8\right)_B,\, f(b_2)=\left(-\frac53,\frac43,-2\right)_B\text{ and }f(b_3)=(5,-4,6)_B.$$So,$$C=\begin{bmatrix}\frac{20}3&-\frac53&5\\-\frac{16}3&\frac43&-4\\8&-2&6\end{bmatrix}.$$
Best Answer
We wish to find a $3\times 3$ matrix $T$ such that $TA=B$ where \begin{align*} A &=\begin{bmatrix}1 & 0 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} & B &= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix} \end{align*} Perhaps the quickest way to find $T$ is to multiply the equation $TA=B$ on the right by $A^{-1}$ to obtain $$ T=BA^{-1} $$ Can you compute $A^{-1}$ and carry out the matrix multiplication?