We have that $\mathcal{B}:=\left (b_1:=\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}, \ b_2:=\begin{pmatrix}1 \\ 0\\ -1\end{pmatrix}, \ b_3:=\begin{pmatrix}-1 \\ 1\\ 0\end{pmatrix}\right )$ is a basis of $\mathbb{R}^3$.
I want to determine the following:
$\gamma_{\mathcal{B}}\left (\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}\right )$
$M_I^{\mathcal{B}}(\text{id})$, $M_{\mathcal{B}}^I(\text{id})$
$M_{\mathcal{B}}^{\mathcal{B}}(\phi_A)$ with $A=\begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$.
These are all matrices for changing the basis, or not?
But how do we calculate that? Do we write the matrix as a linear combination of the vectors of the basis?
I can only guess the meaning of your notations.
Assuming $\gamma_B$ are the coordinates of $(3, -1, 2)^\top$ with respect to the basis $B$, simply solve
$$\begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = \lambda_1b_2 + \lambda_2b_2 +\lambda_3b_3.$$
Let $\operatorname{id}$ be the identity, that is $$\operatorname{id}:\mathbb{R}^3 \to \mathbb{R}^3,\ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \mapsto \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$
Then $M_I^B(\operatorname{id})$ is the transformation matrix from the basis $B$ to the standard basis $I$. The columns of $M_I^B(\operatorname{id})$ are the coordinates of each basis vector $b_i$ with respect to the basis vectors of $I$. The general approach would be to solve the following three equations:
$$b_1 = \lambda_{11}e_1 + \lambda_{21}e_2 + \lambda_{31}e_3$$ $$b_2 = \lambda_{12}e_1 + \lambda_{22}e_2 + \lambda_{32}e_3$$ $$b_3 = \lambda_{13}e_1 + \lambda_{23}e_2 + \lambda_{33}e_3$$
However, since our linear transformation is the identity and since the coordinate vector $$\gamma_I\left[\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\right]$$ of any vector with respect to the standard basis $I$ is precisely the vector itself, your matrix $M_I^B(\operatorname{id})$ is simply the matrix whose columns are exactly the basis vectors of $B$, thus
$$M_I^B(\operatorname{id}) = \begin{pmatrix} \vert & \vert & \vert \\ b_1 & b_2 & b_3 \\ \vert & \vert &\vert \end{pmatrix} $$
Analogously, $M_B^I(\operatorname{id})$ is the matrix whose columns are the coordinate vectors of each standard basis vector $e_i$ with respect to the basis $B$.
$M_B^B(\phi_A)$ is simply the matrix whose columns consist of the coordinate vectors of the image $\phi_A(b_i)$ of each basis vector $b_i$ under the linear transformation $\phi_A$ with respect to the same basis $B$.
Addendum:
$M_B^B(\phi_A)$ is the matrix whose columns consist of the coordinates of $\phi_A(b_i)$ with respect to $B$. This leads to solving the three equations
$$\phi_A(b_i) = \lambda_{1i}b_1+\lambda_{2i}b_2+\lambda_{3i}b_3$$ for $i=1,2,3$ and your solutions $\lambda_{ij}$ are the column entries of your desired matrix $M_B^B(\phi_A)$.
So the first equation $$\phi_A(b_1) = \lambda_{11}b_1+\lambda_{21}b_2+\lambda_{31}b_3$$ gives you the first column etc. Note that for any vector $v \in \mathbb{R}^3$ the vector $\phi_A(v)$ is given by $$\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} v_3 \\ v_1 \\ v_1 \end{pmatrix},$$ therefore $$\phi_A(b_1) = b_1,\ \phi_A(b_2) = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ \phi_A(b_3) = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}.$$