The conclusion of Theorem BNS gives us everything this question asks for. We need the reduced row-echelon form of the matrix so we can determine the number of vectors in $T$, and their entries.
\begin{align*}
\begin{bmatrix}
2 & 1 & 1 & 1 \\
-4 & -3 & 1 & -7 \\
1 & 1 & -1 & 3
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & 2 & -2 \\
0 & \leading{1} & -3 & 5 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
We can build the set $T$ in immediately via Theorem BNS, but we will illustrate its construction in two steps. Since $F=\set{3,\,4}$, we will have two vectors and can distribute strategically placed ones, and many zeros. Then we distribute the negatives of the appropriate entries of the non-pivot columns of the reduced row-echelon matrix.
\begin{align*}
T&=\set{
\colvector{\\\\1\\0},\,
\colvector{\\\\0\\1}
}
&
T&=\set{
\colvector{-2\\3\\1\\0},\,
\colvector{2\\-5\\0\\1}
}
\end{align*}