We have
\begin{align*}
\lt{T}{\vect{e}_1} &= \colvector{1\\0\\1\\1\\3}
&
\lt{T}{\vect{e}_2} &= \colvector{3\\1\\2\\0\\1}
&
\lt{T}{\vect{e}_3} &= \colvector{2\\1\\1\\1\\2}
\end{align*}
Apply Theorem LIVRN to test the linear independence of $T$.
\begin{align*}
\begin{bmatrix}
1 & 3 & 2\\
0 & 1 & 1\\
1 & 2 & 1\\
1 & 0 & 1\\
3 & 1 & 2
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & 0\\
0 & \leading{1} & 0\\
0 & 0 & \leading{1} \\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{align*}
so since $r = 3 = n$, the set of vectors $\set{\lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_2},\,\lt{T}{\vect{e}_3}}$ is linearly independent.