= Solution
The second coordinate is already an <eigenvector> with <eigenvalue> $-2$. On the first and third coordinates the <characteristic polynomial> is
$$
\det\begin{pmatrix}2-\lambda&2\\2&-1-\lambda\end{pmatrix}
=\lambda^2-\lambda-6=(\lambda-3)(\lambda+2).
$$
Thus the full <eigenvalues> are $3,-2,-2$. A corresponding <orthonormal eigenbasis> is
$$
\boxed{\begin{array}{c|c}
\lambda=3 & \mathbf v_3=(2,0,1)/\sqrt5\\
\lambda=-2 & \mathbf v_{-2}=(1,0,-2)/\sqrt5,\quad\mathbf e_2=(0,1,0).
\end{array}}
$$
Their pairwise dot products vanish and each has norm one. Substitution checks $A\mathbf v_3=3\mathbf v_3$ and $A\mathbf v_{-2}=-2\mathbf v_{-2}$. The two-dimensional repeated <eigenspace> permits any orthonormal pair spanning the same plane; the displayed choice is convenient because it separates the second coordinate.
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