Any vector from $\complex{3}$ will lead to a consistent system, and therefore there is no vector that will lead to an inconsistent system.
How do we convince ourselves of this? First, row-reduce $E$,
\begin{equation*}
E\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & 1\\
0 & \leading{1} & 0 & 1\\
0 & 0 & \leading{1} & 1
\end{bmatrix}
\end{equation*}
If we augment $E$ with any vector of constants, and row-reduce the augmented matrix, we will never find a pivot column in the final column, so by Theorem RCLS the system will always be consistent.
Said another way, the column space of $E$ is all of $\complex{3}$, $\csp{E}=\complex{3}$. So by Theorem CSCS any vector of constants will create a consistent system (and none will create an inconsistent system).