Solution:
Given that
$\begin{array}{l}
B=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]_{A}=\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
4 & -1 & 5
\end{array}\right] \\
B A=\left[\begin{array}{ccc}
2+4-0 & 2-2+0 & -4+4+0 \\
-4-12+16 & 4+6-4 & -8-12+20 \\
0-4+8 & 0-2+2 & 0-4+10
\end{array}\right] \\
B A=\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right]
\end{array}$
Now, it can be seen that it is $B A=61$. Where I is the unit Matrix
$\text { Or, } B^{-1}=\frac{1}{6}\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
4 & -1 & 5
\end{array}\right]$
Now we can write the given equation as:
$\begin{array}{l}
{\left[\begin{array}{ccc}
0 & 1 & 2 \\
1 & -1 & 0 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
7 \\
3 \\
17
\end{array}\right]} \\
\mathrm{A} \mathrm{X}=\mathrm{B}
\end{array}$
Or, $X=B^{-1} A$
$\begin{array}{l}
=\frac{1}{6}\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
4 & -1 & 5
\end{array}\right]\left[\begin{array}{c}
7 \\
3 \\
17
\end{array}\right] \\
{\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}
14+6-68 \\
-28+6-68 \\
28-3+85
\end{array}\right]} \\
=\frac{1}{6}\left[\begin{array}{c}
-48 \\
-90 \\
110
\end{array}\right] \\
\quad \mathrm{X}=\left[\begin{array}{c}
-8 \\
-15 \\
\frac{110}{6}
\end{array}\right]
\end{array}$
As a result, $x=-8, y=-15$ and $z=\frac{110}{6}$