Determine a basis for a solution set of a linear system I'm studying, in this class we are given practice problems (but no answer sheet, go figure!). So my question is:

Is my answer correct? Please point out anything I might have gotten wrong. 

Determine a basis for the solution set of the linear system:
$$\left\{\begin{array}{l}
3x + 4y - 2z =0\\
2x + 3y + z =0\\
-2x -2y + 6z =0\\
\end{array}\right.$$

My solution:

Then the augmented matrix is:
\begin{align*}
\left[ \begin{array}{ccc|c}
        3 & 4 & -2 &0\\
        2 & 3 & 1 &0\\
        -2 & -2 & 6 &0\\
        \end{array}
\right] 
\end{align*}
Reduced Row Echelon Form 
\begin{align*}
\left[ \begin{array}{ccc|c}
        1 & 0 & -10 &0\\
        0 & 1 & 7 &0\\
        0 & 0 & 0 &0\\
        \end{array}
\right] 
\end{align*}
This means that any vector $\vec x$ can be represented as 
\begin{align*}
\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = 
\begin{bmatrix}
10z\\
-7z\\
z\end{bmatrix}=z 
\begin{bmatrix}
10\\
-7\\
1
\end{bmatrix}
\end{align*}
Where $z$ is any scalar.
This means that the linear system represents a one-dimensional subspace and its basis is:
\begin{align*}\left(\begin{bmatrix}
10\\
-7\\
1
\end{bmatrix}\right)
\end{align*}

Your computations are correct, and so is your result.
