Find a basis of R^3 defined by the equation,

$\displaystyle 2x_{1}+3x_{2}+x_{3} = 0$

I took the above equation and solved for x3, assuming that was saying that x3 could be created using -2 copies of x1 minus 3 copies of x2...

$\displaystyle 2x_{1}+3x_{2} = -x_{3}$

$\displaystyle -2x_{1}-3x_{2} = x_{3}$

Then I solved this like I solve for a kernel.

$\displaystyle \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}=\begin{bmatrix}t\\ s\\ -2t-3s\end{bmatrix}= t\begin{bmatrix}1\\ 0\\ -2\end{bmatrix}+s\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$

Which gave me a result of,

$\displaystyle \begin{bmatrix}1\\ 0\\ -2\end{bmatrix},\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$

However, the answer is:

$\displaystyle \begin{bmatrix}-3\\ 2\\ 0\end{bmatrix},\begin{bmatrix}-1\\ 0\\ 2\end{bmatrix}$

Where did I go wrong? Any help is extremely appreciated. Thanks!