Let $A$ be the matrix whose columns are the vectors in $T$. Then row-reduce $A$,
\begin{equation*}
A\rref B=
\begin{bmatrix}
\leading{1} & 0 & 0 & 2 \\
0 & \leading{1} & 0 & -1 \\
0 & 0 & \leading{1} & 1
\end{bmatrix}
\end{equation*}
From Theorem BS we can form $R$ by choosing the columns of $A$ that have the same indices as the pivot columns of $B$. Theorem BS also guarantees that $R$ will be linearly independent.
\begin{equation*}
R=\set{
\colvector{1 \\ -1 \\ 2},\,
\colvector{3 \\ 0 \\ 1},\,
\colvector{4 \\ 2 \\ 3}
}
\end{equation*}
That was easy. To find $S$ will require a bit more work. From $B$ we can obtain a solution to $\homosystem{A}$, which by Theorem SLSLC will provide a nontrivial relation of linear dependence on the columns of $A$, which are the vectors in $T$. To wit, choose the free variable $x_4$ to be 1, then $x_1=-2$, $x_2=1$, $x_3=-1$, and so
\begin{equation*}
(-2)\colvector{1 \\ -1 \\ 2}+
(1)\colvector{3 \\ 0 \\ 1}+
(-1)\colvector{4 \\ 2 \\ 3}+
(1)\colvector{3 \\ 0 \\ 6}
=
\colvector{0\\0\\0}
\end{equation*}
this equation can be rewritten with the second vector staying put, and the other three moving to the other side of the equality,
\begin{equation*}
\colvector{3 \\ 0 \\ 1}
=
(2)\colvector{1 \\ -1 \\ 2}+
(1)\colvector{4 \\ 2 \\ 3}+
(-1)\colvector{3 \\ 0 \\ 6}
\end{equation*}
We could have chosen other vectors to stay put, but may have then needed to divide by a nonzero scalar. This equation is enough to conclude that the second vector in $T$ is “surplus” and can be replaced (see the careful argument in Example RSC5). So set
\begin{equation*}
S=\set{
\colvector{1 \\ -1 \\ 2},\,
\colvector{4 \\ 2 \\ 3},\,
\colvector{3 \\ 0 \\ 6}
}
\end{equation*}
and then $\spn{S}=\spn{T}$. $T$ is also a linearly independent set, which we can show directly. Make a matrix $C$ whose columns are the vectors in $S$. Row-reduce $C$ and you will obtain the identity matrix $I_3$. By Theorem LIVRN, the set $S$ is linearly independent.