find matrix $A$ in other basis Original matrix is:
$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$
original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$
basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$
Where:
$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} \\f_3=\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3}$$
Therefore:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1f_1+y_2f_2+y_3f_3$$
Expanding that we get:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1(2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3})+y_2(3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} )+y_3(\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3})$$
How should I group things at right side around unit vectors? for example for first coordinate $2\mathbf{e_1} \, !=\, 3\mathbf{e_1}$, how could I group coordinates?

${\Large \textbf {Solution:}}$
Opening brackets:
$$x\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3} = 2y_1\mathbf{e_1}+3y_1\mathbf{e_2}+y_1\mathbf{e_3}+3y_2\mathbf{e_1}+4y_2\mathbf{e_2}+y_2\mathbf{e_3}+y_3\mathbf{e_1}+2y_3\mathbf{e_2}+2y_3\mathbf{e_3}$$
Sorting by unit vectors:
$$x_1\mathbf{e_2}+x_2\mathbf{e_3}+x_3\mathbf{e_3} = (2y_1+3y_2+y_3)\mathbf{e_1}+(3y_1+4y_2+2y_3)\mathbf{e_2}+(y_1+y_2+2y_3)\mathbf{e_3} \tag{1}$$
Based on fotmula:
$$A'=T^{-1}AT$$
Matrix $T$ is based on right side of equation $(1)$, and we have:
$$T=\begin{pmatrix} 2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$
The rest is more less technical.
$$T^{-1}=\begin{pmatrix} -6&5&-2\\4&-3&1\\1&-1&1 \end{pmatrix}$$

Call the first and second basis $E$ and $F$ respectively. You are given $$A_E=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$ and the change of coordinates matrix that changes $F$ coordinates to $E$ coordinates is $$Q=\begin{pmatrix}2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$ Then $$A_F=Q^{-1}A_EQ.$$
