A linear transformation that is not injective will have a nontrivial kernel (Theorem KILT), and this is the key to finding the desired inputs. We need one nontrivial element of the kernel, so suppose that $\vect{z}\in\complex{4}$ is an element of the kernel,
\begin{equation*}
\colvector{0\\0\\0}
=\zerovector
=\lt{T}{\vect{z}}
=\colvector{
2z_1+z_2+z_3\\
-z_1+3z_2+z_3-z_4\\
3z_1+z_2+2z_3-2z_4
}
\end{equation*}
Vector equality Definition CVE leads to the homogeneous system of three equations in four variables,
\begin{align*}
2z_1+z_2+z_3&=0\\
-z_1+3z_2+z_3-z_4&=0\\
3z_1+z_2+2z_3-2z_4&=0
\end{align*}
The coefficient matrix of this system row-reduces as
\begin{equation*}
\begin{bmatrix}
2 & 1 & 1 & 0 \\
-1 & 3 & 1 & -1 \\
3 & 1 & 2 & -2
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & 1 \\
0 & \leading{1} & 0 & 1 \\
0 & 0 & \leading{1} & -3
\end{bmatrix}
\end{equation*}
From this we can find a solution (we only need one), that is an element of $\krn{T}$,
\begin{equation*}
\vect{z}=\colvector{-1\\-1\\3\\1}
\end{equation*}
Now, we choose a vector $\vect{x}$ at random and set $\vect{y}=\vect{x}+\vect{z}$,
\begin{align*}
\vect{x}
&=\colvector{2\\3\\4\\-2}
&
\vect{y}&=\vect{x}+\vect{z}=
\colvector{2\\3\\4\\-2}+\colvector{-1\\-1\\3\\1}
=\colvector{1\\2\\7\\-1}
\end{align*}
and you can check that
\begin{equation*}
\lt{T}{\vect{x}}
=\colvector{11\\13\\21}
=\lt{T}{\vect{y}}
\end{equation*}
A quicker solution is to take two elements of the kernel (in this case, scalar multiples of $\vect{z}$) which both get sent to $\zerovector$ by $T$. Quicker yet, take $\zerovector$ and $\vect{z}$ as $\vect{x}$ and $\vect{y}$, which also both get sent to $\zerovector$ by $T$.