= Solution
The first coordinate direction is already an <eigenvector> of $A$ with <eigenvalue> $4$. On the last two coordinates, the <matrix> is $\begin{pmatrix}2&4\\4&2\end{pmatrix}$, with <characteristic polynomial>
$$
\det\begin{pmatrix}\lambda-2&-4\\-4&\lambda-2\end{pmatrix}=(\lambda-2)^2-16=(\lambda-6)(\lambda+2).
$$
Thus the three <eigenvalues> are $4,6,-2$. For $\lambda=6$, the last two components are equal; for $\lambda=-2$, they are opposite. Normalising by the <Euclidean norm> gives the complete list
$$
\boxed{\begin{array}{c|c}
\lambda&\text{unit eigenvector}\\\hline
4&(1,0,0)^T\\
6&(0,1,1)^T/\sqrt2\\
-2&(0,1,-1)^T/\sqrt2
\end{array}}
$$
Each sign-reversed <vector> is an equally valid choice. The three displayed <eigenvectors> are mutually <orthogonal>, as expected for a real <symmetric matrix> with distinct <eigenvalues>.
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