Determining if a vector is in the row space

I am just trying to determine if the vector $[0, 7, 4]$ belongs in the row space of the matrix $$A = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 1 & -5 & 4 \\ \end{bmatrix}$$ What I have done so far is created an augmented matrix like so $$\left[ \begin{array}{ccc|c} 1 & 3 & 1 & 0 \\ 2 & -1 & -5 & 7 \\ 0 & 4 & 4 & 4 \\ \end{array} \right]$$ (naive gaussian) reducing to $$\left[ \begin{array}{ccc|c} 1 & 3 & 1 & 0 \\ 0 & -7 & -7 & 7 \\ 0 & 0 & 0 & 32/7 \\ \end{array} \right]$$ returning inconsistent, i.e. not existing in the row space. However apparently it does in fact belong in the row space, so clearly I have gone about this the wrong way. Is someone able to correct this?

A vector $\vec b$ is in the row space of $A$ if and only if $\vec b$ is in the column space of $A^\top$. Thus, to determine if the vector $\vec b=\left[\begin{array}{r} 0 \\ 7 \\ 4 \end{array}\right]$ is in the row space of $A = \left[\begin{array}{rrr} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 1 & -5 & 4 \end{array}\right]$, form the augmented matrix $$\left[\begin{array}{rrr|r} 1 & 3 & 1 & 0 \\ 2 & -1 & -5 & 7 \\ 0 & 4 & 4 & 4 \end{array}\right]$$ Row reducing gives $$\DeclareMathOperator{rref}{rref}\rref\left[\begin{array}{rrr|r} 1 & 3 & 1 & 0 \\ 2 & -1 & -5 & 7 \\ 0 & 4 & 4 & 4 \end{array}\right]= \left[\begin{array}{rrr|r} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$ This gives an inconsistent system. Hence $\vec b$ is not in the row space of $A$.

Consider the matrix$$\begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 1 & -5 & 4 \\ \end{bmatrix}$$
Minus $2$ times of row $1$ plus row $2$ gives us row $3$, the row space is spanned by the first two rows.
Suppose $$a\begin{bmatrix} 1 & 2 & 0 \end{bmatrix} + b\begin{bmatrix} 3 & -1 & 4 \end{bmatrix}=\begin{bmatrix} 0 & 7 & 4 \end{bmatrix}$$
From the third coordinate, $b=1$.
Hence $a+3=0$ and $2a-1=7$ of which we can see a contradiction.