Answer
See below
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
-3\\
9\\
5
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
e^{2t}\\
3e^{2t}\\
e^{2t}
\end{bmatrix}$
$x_3(t)=\begin{bmatrix}
e^{4t}\\
e^{4t}\\
e^{4t}
\end{bmatrix}$
$A=\begin{bmatrix}
2 & -1 & 3\\
3 & 1 & 0\\
2 & -1 & 3
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
2e^{2t}\\
6e^{2t}\\
2e^{2t}
\end{bmatrix}$
$x_3(t)=\begin{bmatrix}
4e^{4t}\\
4e^{4t}\\
4e^{4t}
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
2 & -1 & 3\\
3 & 1 & 0\\
2 & -1 & 3
\end{bmatrix}\begin{bmatrix}
-3\\
9\\
5
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}=x_1'(t)$
$Ax_2(t)=\begin{bmatrix}
2 & -1 & 3\\
3 & 1 & 0\\
2 & -1 & 3
\end{bmatrix}\begin{bmatrix}
e^{2t}\\
3e^{2t}\\
e^{2t}
\end{bmatrix}=\begin{bmatrix}
2e^{2t}\\
6e^{2t}\\
2e^{2t}
\end{bmatrix}=x_2'(t)$
$Ax_3(t)=\begin{bmatrix}
2 & -1 & 3\\
3 & 1 & 0\\
2 & -1 & 3
\end{bmatrix}\begin{bmatrix}
e^{4t}\\
e^{4t}\\
e^{4t}
\end{bmatrix}=\begin{bmatrix}
4e^{4t}\\
4e^{4t}\\
4e^{4t}
\end{bmatrix}=x_3'(t)$
The Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
-3 & e^{2t} & e^{4t}\\
9 & 3e^{2t} & e^{4t}\\
5 & e^{2t} & e^{4t}
\end{vmatrix}=-16e^{6t}$
Since the Wronskian is never zero, it follows that $\{x_1, x_2\}$ is linearly independent on any interval and so is a fundamental set of solutions for the given vector differential equation. Therefore, the general solution to the system is
$x(t)=\begin{bmatrix}
-3c_1 +c_2e^{2t}+c_3e^{4t} \\
9c_1+3c_2e^{2t}+c_3e^{4t}\\
5c_1+c_2e^{2t}+c_2e^{4t}
\end{bmatrix}$
