Answer
$\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]$ is an eigenvector of $A$ for the eigenvalue $-2$.
Work Step by Step
Given $$ A=\left[\begin{array}{lll}
3 & 6 & 7 \\
3 & 3 & 7 \\
5 & 6 & 5
\end{array}\right],\ \ \ \ \ \mathbf{v}=\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]$$
Since
\begin{align*}
A \mathbf{v}&=\left[\begin{array}{lll}
3 & 6 & 7 \\
3 & 3 & 7 \\
5 & 6 & 5
\end{array}\right]\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\\
&=\left[\begin{array}{c}
-2 \\
4 \\
-2
\end{array}\right]=(-2)\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]
\end{align*}
then $A \mathbf{v}=0 \mathbf{v}$, hence $\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]$ is an eigenvector of $A$ for the eigenvalue $-2$.
