Answer
$\lambda =4$ is eigenvalue
$\mathbf{v}= \left[\begin{array}{r}
-1\\
-1\\
1
\end{array}\right]$
Work Step by Step
Given $$A= \left[\begin{array}{rrr}
3 & 0 & -1 \\
2 & 3 & 1 \\
-3 & 4 & 5
\end{array}\right]$$
Since
\begin{align*}
|A-\lambda I|&= 0\\
\left|\begin{array}{rrr}
3 -\lambda& 0 & -1 \\
2 & 3-\lambda & 1 \\
-3 & 4 & 5-\lambda
\end{array}\right|&= 0\\
-λ^3+11λ^2-32λ+16&=0\\
\end{align*}
Since $λ=4$ satisfies the characteristic equation , then $λ=4$ is an eigenvalue
and the corresponding eigenvector given by
\begin{align*}
\left[\begin{array}{rrr}
3-4 & 0 & -1 \\
2 & 3-4 & 1 \\
-3 & 4 & 5-4
\end{array}\right]\left[\begin{array}{r}
x_1\\
x_2\\
x_3
\end{array}\right]&=\left[\begin{array}{r}
0\\
0\\
0
\end{array}\right]\\
\end{align*}
then
$$ \left[\begin{array}{rrr}
-1 & 0 & -1 \\
2 & -1 & 1 \\
-3 & 4 & 1
\end{array}\right] \to \left[\begin{array}{rrr}
-1 & 0 & -1 \\
0 & -1 & -1 \\
0 & 4 & 4
\end{array}\right] \to \left[\begin{array}{rrr}
-1 & 0 & -1 \\
0 & -1 & -1 \\
0 & 0 & 0
\end{array}\right]$$
then $x_3= t,\ \ x_2=-t,\ \ x_1 = -t $, then the eigenvector is $ \left[\begin{array}{r}
-1\\
-1\\
1
\end{array}\right]$
