Write $X$ as a collection of columns, ie, $X = \begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}$. Then $AX = cX$ iff $Ax_i = c x_i$ for all $i$ iff $x_i \in \ker (A-cI)$ for all $i$.
In this particular case, since $5$ is the only eigenvalue, $\ker (A-cI) = \{0\}$ if $c\neq 5$ and $\ker (A-5I) = \mathbb{sp} \{e_3 \}$.
So all solutions are of the form $c \neq 5, X = 0$, or $c=5$ with $X= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \alpha_1 & \alpha_2 & \alpha_3 \end{bmatrix}$, for arbitrary constants $\alpha_i$.
Addendum: Here is an approach that doesn't use the eigenvalue/eigenvector language:
As above, write $X$ as a collection of columns. Then we still have $AX = cX$ iff $Ax_i = c x_i$ for all $i$, so we are interested in looking for solutions to the equation$Ax = cx$ or equivalently $(A-cI) x = 0$.
There are two cases to consider. If $c \neq 5$, the the matrix $A-cI$ is invertible, since it is lower triangular with non-zero diagonal elements. Consequently, if $(A-cI)x = 0$, then we must have $x=0$ (by multipliing both sides by the matrix $(A-cI)^{-1}$, or noting that the first row implies $x_1=0$, the second row then implies $x_2 = 0$ and finally the third row gives $x_3=0$).
Now consider the case $c = 5$, and consider $(A-5I)x = 0$. Notice that the second row implies $x_1=0$ and the third row implies $x_2=0$. Furthermore, $x_3$ can have any value whatsoever.
Combining the above, we see that $X$ must have the form $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \alpha_1 & \alpha_2 & \alpha_3 \end{bmatrix}$, for arbitrary constants $\alpha_i$.