Change of Basis over $\mathbb{Z}$

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$[v ]_B$ coordinate vector of v with respect to $B$
$[v ]_C$ coordinate vector of v with respect to $C$
${}_C[ Id_V ]_B$ change of basis matrix from $B$ to $C$

For example we have the following set of vectors in $\mathbb{Z_3^3}$
$B$ = {[1 0 2]${^T}$, [2 1 1]${^T}$ , [1 1 1]${^T}$}
$C$ = {[1 1 1]${^T}$ ,[2 1 1]${^T}$ , [1 1 2]${^T}$}
How can I compute the change of basis from $B$ to $C$ using the following formula:
${}_C[ Id_V ]_B [v ]_B = [ v ]_C$ where ,
${}_C[ Id_V ]_B = [ [b_1]_C \cdots [b_n]_C]$
Again $V$ belongs to $\mathbb{Z_3^3}$ and $Id_V : V \rightarrow V$

I think the formula is a fairly simple computation but the confusing part is the change of coordinate vectors , namely ${}_C[ Id_V ]_B = [ [b_1]_C \cdots [b_n]_C]$. Also, I have checked that the given set of vector do form basis of $\mathbb{Z_3^3}$

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What a strange notation.

Note that ${}_I[Id_V]_B=B$, where $I$ is the unit matrix, and then ${}_C[Id_V]_I=C^{-1}$ and ${}_C[Id_V]_B = {}_C[Id_V]_I \, {}_I[Id_V]_B=C^{-1}B$.

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${}_C[Id_V]_B = {}_C[Id_V]_I \, {}_I[Id_V]_B=C^{-1}B$

= $\begin{bmatrix} 2 \:0\: 2 \\ 1 \:2\: 0 \\ 0 \:2\: 1\end{bmatrix}$ $\begin{bmatrix} 1 \:2\: 1 \\ 0 \:1\: 1 \\ 2 \:1\: 1\end{bmatrix}$

=$\begin{bmatrix} 0 \:0\: 1 \\ 1 \:1\: 0 \\ 2 \:0\: 0\end{bmatrix}$