Answer
$λ=3$ is an eigenvalue of the given matrix
$\mathbf{v} = \left[\begin{array}{r}
3\\2\\1
\end{array}\right] $
Work Step by Step
Given $$\left[\begin{array}{rrr}
1 & 2 & 2 \\
3 & -2 & 1 \\
0 & 1 & 1
\end{array}\right] $$
Since
\begin{align*}
|A-\lambda I |&= 0\\
\left|\begin{array}{rrr}
1-\lambda & 2 & 2 \\
3 & -2-\lambda & 1 \\
0 & 1 & 1-\lambda
\end{array}\right|&=0\\
\left(1-λ\right)\left(λ^2+λ-3\right)-2\cdot \:3\left(-λ+1\right)+2\cdot \:3&=0\\
-λ^3+10λ-3&=0
\end{align*}
Since $ λ=3$ satisfies the characteristic equation , then $λ=3$ is an eigenvalue of the given matrix , to find the corresponding eigenvector
\begin{align*}
(A-\lambda I) \mathbf{v}&=0\\
\left[\begin{array}{rrr}
-2 & 2 & 2 \\
3 & -5 & 1 \\
0 & 1 & -2
\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
$$\left[\begin{array}{rrr}
-2 & 2 & 2 \\
3 & -5 & 1 \\
0 & 1 & -2
\end{array}\right]\to \left[\begin{array}{rrr}
-1 & 1 & 1 \\
3 & -5 & 1 \\
0 & 1 & -2
\end{array}\right]\to \left[\begin{array}{rrr}
-1 & 1 & 1 \\
0& -2 &4 \\
0 & 1 & -2
\end{array}\right] \to \left[\begin{array}{rrr}
-1 & 1 & 1 \\
0& -2 &4 \\
0 & 0 & 0
\end{array}\right]$$
Then $ x_3= t,\ \ \ x_2= 2t,\ \ \ x_1= 3t$, hence $\mathbf{v} = \left[\begin{array}{r}
3\\2\\1
\end{array}\right] $
