Answer
$\vec{x}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
Work Step by Step
Since the third column of the matrix is the sum of the first column and two times the second column, we have the following:
$ \begin{bmatrix}4&1&6\\-7&5&3\\9&-3&3\end{bmatrix}\begin{bmatrix}1\\2\\-1\end{bmatrix}=1\cdot\begin{bmatrix}4\\-7\\9\end{bmatrix}+2\cdot\begin{bmatrix}1\\5\\-3\end{bmatrix}+(-1)\cdot\begin{bmatrix}6\\3\\3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
Any nonzero constant multiple of this solution will also work as a nontrivial solution to the homogeneous equation.
