<s>[INST] Find a basis for a solution set of a linear system I'm trying hard with this exercise, but it is breaking my back.

Find a basis for the solution set of the given homogeneous linear system
$3x_1+x_2+x_3=0$
  $6x_1+2x_2+2x_3=0$
  $-9x_1-3x_2-3x_3=0$

I do what I know I need to do. First I get the solution set of the system by reducing like this:
$\begin{pmatrix}
3 & 1 & 1 \\
6 & 2 & 2 \\
-9 & -3 & -3 \end{pmatrix} \leadsto \begin{pmatrix}
3 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \leadsto\begin{pmatrix}
1 & 1/3 & 1/3 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}$
So I know $\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix}$
That being the general solution. 
Now, giving the values for $r$ and $s$ according to the standard vectors $i$, $j$
$\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix} = r \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix} + s\begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix}$
From my results, the basis will be: 
$ ( \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix}, \begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix})$
But instead, the book answer (I'm self-studying )is:
$ ( \begin{bmatrix}
-1\\
3\\
0\end{bmatrix}, \begin{bmatrix}
-1\\
0\\
3\end{bmatrix})$
Any idea on what I'm doing wrong? Thank you :)
 [/INST] Check your expression for $\vec x$.  I think it should be $\vec x=\begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=\begin{bmatrix} -\frac{1}{3}r-\frac{1}{3}s \\r \\s \end{bmatrix}=r\begin{bmatrix} -\frac{1}{3} \\1 \\0 \end{bmatrix}+s\begin{bmatrix} -\frac{1}{3} \\0 \\1 \end{bmatrix}$
</s>