First, find the eigenvalues of the matrix A.
\begin{equation*}
\left| \begin{matrix} 1-\lambda & 0 & 0 \\2 & 1-\lambda & -2 \\ 3& 2 & 1-\lambda\end{matrix}\right| = 0 \notag
\end{equation*}
\begin{gather}
(1 - \lambda)( (1-\lambda)(1-\lambda) + 4) = 0,\label{eq-1} \\
(1 - \lambda)( \lambda^2 - 2\lambda + 5) = 0. \label{eq-2}
\end{gather}
The eigenvalues are $ \lambda = 1, \lambda = 1 + 2i,\ \lambda = 1- 2i$.
When $ \lambda = 1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
0 & 0 & 0
\\2 & 0 & -2
\\ 3 & 2 & 0\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 2 & 3
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(2, -3, 2)^T$ and corresponding solution is $x^{(1)} = c_1e^t(2, -3, 2)^T $.
When $ \lambda = 1 + 2i$,
\begin{equation*}
A-\lambda I=\begin{bmatrix}
-2i & 0 & 0
\\2 & -2i & -2
\\ 3 & 2 & -2i\end{bmatrix}\cong
\begin{bmatrix}
1 & 0 & 0
\\0 & i & 1
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(0, i, 1) ^T$ and corresponding solution is
\begin{equation*}
(0, i, 1)^T e^{(1+2i)t} = (0, i, 1)^T (e^t)( e^{2it}) = (0, i, 1)^T (e^t)(\cos 2t + i\sin 2t) = (0, i\cos 2t - \sin 2t, \cos 2t + i\sin 2t)^T (e^t) $
=(e^t)[ (0, -\sin 2t, \cos 2t)^T +i(0, \cos 2t, \sin 2t)^T]
\end{equation*}
The general solution is
$x = e^t\left[ c_1 (2, -3, 2)^T + c_2 (0, -\sin 2t, \cos 2t)^T +c_3 (0, \cos 2t, \sin 2t)^T\right]$
