Answer
$ \begin{bmatrix}1\\1\\-1\end{bmatrix} $
Work Step by Step
Since the third column of the matrix is the sum of the first two, we have the following:
$ \begin{bmatrix}2&3&5\\-5&1&-4\\-3&-1&-4\\1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix}=1\cdot\begin{bmatrix}2\\-5\\-3\\1\end{bmatrix}+1\cdot\begin{bmatrix}3\\1\\-1\\0\end{bmatrix}+(-1)\cdot\begin{bmatrix}5\\-4\\-4\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix} $.
In fact, any vector of the form $(c,c,-c)$, $c\neq 0$, is a nontrivial solution to the given homogeneous equation.
