Answer
Yes, this $\vec{b}$ is in the range of this linear transformation.
Work Step by Step
We simply row reduce the augmented matrix $[\mathbf{A}\ \ \vec{b}]$ and find that there are no rows of the form $[0\ 0\ 0\ 0\ b]$ with $b\neq 0$. Hence, the equation $T(\vec{x})=\mathbf{A}\vec{x}=\vec{b}$ is consistent.
You should find that the equation has infinitely many solutions of the form
$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}=\begin{bmatrix}3+9s-7t\\1+4s-3t\\s\\t\end{bmatrix}$
