Can you solve it using the matrix form

Swap $R_1$ and $R_2$ to get a leading 1 in the first row:

$$R_1 \leftrightarrow R_2 \\ \begin{bmatrix} 1 & -1 & 2 & | & 4 \\ 2 & 2 & -1 & | & -7 \\ 3 & 3 & -1 & | & 0 \end{bmatrix}$$

Make the elements below the leading 1 in the first column zero:

$$R_2 - 2R_1 \to R_2 \\ R_3 - 3R_1 \to R_3 \\ \begin{bmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 4 & -5 & | & -15 \\ 0 & 6 & -7 & | & -12 \end{bmatrix}$$

Get a leading 1 in the second row, second column:

$$\frac{1}{4}R_2 \to R_2 \\ \begin{bmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 1 & -\frac{5}{4} & | & -\frac{15}{4} \\ 0 & 6 & -7 & | & -12 \end{bmatrix}$$

Make the element below the leading 1 in the second column zero:

$$R_3 - 6R_2 \to R_3 \\ \begin{bmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 1 & -\frac{5}{4} & | & -\frac{15}{4} \\ 0 & 0 & \frac{1}{2} & | & \frac{21}{2} \end{bmatrix}$$

Get a leading 1 in the third row, third column:

$$2R_3 \to R_3 \\ \begin{bmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 1 & -\frac{5}{4} & | & -\frac{15}{4} \\ 0 & 0 & 1 & | & 21 \end{bmatrix}$$

Make the elements above the leading 1 in the third column zero:

$$R_2 + \frac{5}{4}R_3 \to R_2 \\ R_1 - 2R_3 \to R_1 \\ \begin{bmatrix} 1 & -1 & 0 & | & -38 \\ 0 & 1 & 0 & | & \frac{45}{2} \\ 0 & 0 & 1 & | & 21 \end{bmatrix}$$

Make the element above the leading 1 in the second column zero:

$$R_1 + R_2 \to R_1 \\ \begin{bmatrix} 1 & 0 & 0 & | & -\frac{31}{2} \\ 0 & 1 & 0 & | & \frac{45}{2} \\ 0 & 0 & 1 & | & 21 \end{bmatrix}$$

From the matrix, we can directly read the solution:

$$x = -\frac{31}{2}$$ $$y = \frac{45}{2}$$ $$z = 21$$ $$x = -\frac{31}{2}, y = \frac{45}{2}, z = 21$$