Answer
$$x=-{\frac {23}{18}},\quad y= \frac {17}{9}, \quad z=-{\frac {17}{18}}.$$
Work Step by Step
We have the system
$$\left[ \begin {array}{ccc} 0&1&2\\ 3&2&1
\\ 4&-3&-4\end {array} \right]
\left[\begin{array}{rrr}{x}\\ {y} \\{z} \end{array}\right]=\left[\begin{array}{rrr}{0} \\ {-1}\\{-7} \end{array}\right].$$
The coefficients matrix $A= \left[ \begin {array}{ccc} 0&1&2\\ 3&2&1
\\ 4&-3&-4\end {array} \right]
$ has the inverse
$$A^{-1}=\left[ \begin {array}{ccc} {\frac {5}{18}}&\frac{1}{9}&\frac{1}{6}
\\ -{\frac {8}{9}}&\frac{4}{9}&-\frac{1}{3}\\ {
\frac {17}{18}}&-\frac{2}{9}&\frac{1}{6}\end {array} \right]
.
$$
The system has the solution
$$\left[\begin{array}{rrr}{x}\\ {y} \\{z} \end{array}\right]=\left[ \begin {array}{ccc} {\frac {5}{18}}&\frac{1}{9}&\frac{1}{6}
\\ -{\frac {8}{9}}&\frac{4}{9}&-\frac{1}{3}\\ {
\frac {17}{18}}&-\frac{2}{9}&\frac{1}{6}\end {array} \right]\left[\begin{array}{rrr}{0} \\ {-1}\\{-7} \end{array}\right]=\left[ \begin {array}{c} -{\frac {23}{18}}\\ {
\frac {17}{9}}\\ -{\frac {17}{18}}\end {array}
\right]
.$$
That is
$$x=-{\frac {23}{18}},\quad y= \frac {17}{9}, \quad z=-{\frac {17}{18}}.$$
