= Solution
Call the given <matrix> $M$. Its middle coordinate decouples, immediately giving the <eigenvalue> $3$ with normalized <eigenvector> $(0,1,0)^T$. The remaining <characteristic polynomial> comes from the $xz$ block:
$$
\det\begin{pmatrix}-1-\lambda&-3\sqrt3\\-3\sqrt3&5-\lambda\end{pmatrix}=(-1-\lambda)(5-\lambda)-27=\lambda^2-4\lambda-32=(\lambda+4)(\lambda-8).
$$
For $\lambda=-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt3z$; for $\lambda=8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt3x$. After <vector normalization>, one possible complete set is
$$
\boxed{\begin{aligned}
\lambda_1&=-4,&v_1&=\frac12(\sqrt3,0,1)^T,\\
\lambda_2&=3,&v_2&=(0,1,0)^T,\\
\lambda_3&=8,&v_3&=\frac12(-1,0,\sqrt3)^T.
\end{aligned}}
$$
Each has unit <Euclidean norm>. The <dot products> with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt3+\sqrt3)/4=0$. This explicitly verifies <orthonormality>, also predicted by the <spectral theorem> for real <symmetric matrices> with distinct <eigenvalues>. Changing any <eigenvector>'s sign gives an equally valid normalized choice.
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