To find the kernel, we require all $\vect{x}\in\complex{3}$ such that $\lt{T}{\vect{x}}=\zerovector$. This condition is
\begin{equation*}
\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}=\colvector{0\\0}
\end{equation*}
This leads to a homogeneous system of two linear equations in three variables, whose coefficient matrix row-reduces to
\begin{equation*}
\begin{bmatrix}
\leading{1} & -\frac{1}{2} & \frac{5}{2}\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
With two free variables Theorem BNS yields the basis for the null space
\begin{equation*}
\set{
\colvector{-\frac{5}{2}\\0\\1},\,
\colvector{\frac{1}{2}\\1\\0}
}
\end{equation*}
With $\nullity{T}\neq 0$, $\krn{T}\neq\set{\zerovector}$, so Theorem KILT says $T$ is not injective.