= Solution
The matrix is real symmetric, so the <spectral theorem> guarantees an <orthonormal basis> of <eigenvectors>. Its characteristic determinant factors as
$$
\det(A-\lambda I)=-(\lambda+3)(\lambda-6)(\lambda-1).
$$
Solving the three corresponding equations for the <null space> gives the normalized <eigenvectors>
$$
\boxed{\lambda=6:\ \frac1{\sqrt5}\begin{pmatrix}2\\0\\-1\end{pmatrix},\qquad\lambda=-3:\ \begin{pmatrix}0\\1\\0\end{pmatrix},\qquad\lambda=1:\ \frac1{\sqrt5}\begin{pmatrix}1\\0\\2\end{pmatrix}.}
$$
Their pairwise <inner products> vanish. With these vectors as columns,
$$
Q=\begin{pmatrix}2/\sqrt5&0&1/\sqrt5\\0&1&0\\-1/\sqrt5&0&2/\sqrt5\end{pmatrix},\qquad\boxed{Q^TAQ=\operatorname{diag}(6,-3,1).}
$$
Since $Q^TQ=I$, this also gives $A=Q\operatorname{diag}(6,-3,1)Q^T$.
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