Note that
$$A_{BC} = \begin{bmatrix}
1 & 2 & 1 \\
2 & 1 & 0 \\
1 & 1 & 0
\end{bmatrix}$$
Is the change of basis matrix from $C$ to $B$ where $C=\{(1,0,0),\;(0,1,0),\;(0,0,1)\}$ and $B=\{(1,2,1),\;(2,1,1),\;(1,0,0)\}$. Then the matrix you are looking for is
$A_{BC}\times A_B \times (A_{BC})^{-1} $
Please, take a look to this question.
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.
Best Answer
To find the matrix representing a linear transformation in a given basis, apply the linear transformation to each basis vector in turn and write the result as a linear combination of the basis vectors. The coefficients in that linear combination form a column of the matr8ix.
Here, the first basis vector is $\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}$. A applied to that is $\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}$$\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}= \begin{bmatrix}1 \\ 3 \\ 2 \end{bmatrix}$$= 0\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}- 2\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ 3\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}$.
So the first column of the new matrix is $\begin{bmatrix} 0 \\ -2 \\ 3\end{bmatrix}$.