The symmetric matrix (call is $A$) has two eigenvalues, one of multiplicity 2 at -1, and one of multiplicity 1 at 5. The eigenspaces $\ker(A+I)$ and $\ker (A-5I)$ are orthogonal complements, so the only issue is choosing a basis for $\ker(A+I)$ that is orthogonal.
Choose $\frac{1}{\sqrt{3}}(1,1,1)^T$ as a basis for $\ker(A-5I)$ (not a huge amount of choice there).
Since $\ker(A+I) = \ker (1,1,1)^T$, we can choose a element of the null space, say $\frac{1}{\sqrt{2}} (1,-1,0)^T$, and just find an orthogonal vector (in $\ker(A+I)$, of course), for example: $\frac{1}{\sqrt{6}} (1,1,-2)^T$.
It is easy to check that these vectors are orthogonal, in fact, orthonormal.