matrix and linear transformations With regards to this (Matrix associated to a linear transformation with respect to a given basis)
I saw Ivo Terek solution which goes:
\begin{align}L(1,1,1) &= (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
 2\cdot (1,0,0) \\ L(1,1,0) &= (0,8,2) = 2\cdot (1,1,1) + 6\cdot (1,1,0) -
 8\cdot (1,0,0) \\ L(1,0,0) &= (0,3,1) = 1\cdot (1,1,1) + 2\cdot (1,1,0) -
 3\cdot (1,0,0),\end{align}
However, I was not able to understand how:
$ 6\cdot (1,1,1) + 0\cdot (1,1,0) - 2\cdot (1,0,0) \\2\cdot (1,1,1) + 6\cdot (1,1,0) - 8\cdot (1,0,0) \\1\cdot (1,1,1) + 2\cdot (1,1,0) - 3\cdot (1,0,0) $
was derived. I only know half of where the equation came from.

I suppose you understand the first equality in
$$L(1,1,1) = (4,6,6) = 6\cdot (1,1,1) + 0\cdot (1,1,0) -
 2\cdot (1,0,0)$$
well, as $L$ is defined by $L \left( \begin{bmatrix}x_{1}\\x_{2}\\x_{3} \end{bmatrix} \right) = \begin{bmatrix}4x_{3}\\3x_{1}+5x_{2}-2x_{3}\\x_{1}+x_{2}+4x_{3} \end{bmatrix}$ in the linked question.
Observe that for each vector $v$ expressed as a linear combination of basic vectors in the basis $B = \left(\begin{bmatrix}1\\1\\1 \end{bmatrix} , \begin{bmatrix}1\\1\\0 \end{bmatrix}, \begin{bmatrix}1\\0\\0 \end{bmatrix} \right)$, the first basic vector $(1,1,1)^T$ determines the third component of $v$, the second basic vector $(1,1,0)^T$ determines the second component of $v$.
In particular, you have $(4,6,6)^T$, so you need $6$ as the coefficient for $(1,1,1)^T$, as the basic vectors $(1,1,0)^T$ and $(1,0,0)^T$ can't contribute to the third component $z$ in the vector $(x,y,z)^T$.  Use a similar reasoning to get $0$ for the coefficient of $(1,1,0)^T$.  The rest is a simple arithmetic exercise.
