Can't orthogonally diagonalise this symmetric matrix.

If you know there are only eigenvalues $-2$ and $7$, then you can eliminate those with eigenvalue $-2$ and be left with the others. This is done by considering $$ (A+2I) = \left[\begin{array}{ccc} 4 & 2 & -4 \\ 2 & 1 & -2 \\ -4 & -2 & 4\end{array}\right] $$ The columns have to be eigevectors with eigenvalue $7$. There is only one independent column, which is what you expect: $$ \left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right],\;\;\;\; A \left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right] = \left[\begin{array}{c}14 \\ 7 \\ -14\end{array}\right] = 7\left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right] $$

Now consider the case for eigenvalue $-2$: $$ (A-7I) = \left[\begin{array}{ccc} -5 & 2 & -4 \\ 2 & -8 & -2 \\ -4 & -2 & -5\end{array}\right] $$ The column space is two-dimensional; it's obviously not one-dimensional, and you can easily check that the columns are orthogonal to the eigenvector with eigenvalue $7$ given above. So everything checks. If you add the first and the third columns, you find that the following is an eigenvector with eigenvalue $-2$: $$ \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right], $$ and this one is orthogonal to the middle column. So an orthogonal basis of the eigenspace with eigenvalue $-2$ is $$ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array},\right] \left[\begin{array}{c} 1 \\ -4 \\ -1 \end{array}\right]\right\} $$ If you normalize the column vectors, then you get the required orthogonal change of basis matrix $U$ such that $U^{-1}AU=D=\left[\begin{array}{ccc}7 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$ is a diagonal matrix. The matrix $U$ is $$ U = \left[\begin{array}{ccc} 2/3 & 1/\sqrt{2} & 1/\sqrt{18} \\ 1/3 & 0 & -4/\sqrt{18} \\ -2/3 & 1/\sqrt{2} & -1/\sqrt{18} \end{array}\right] $$ The matrix $U$ is orthogonal, which means $U^{-1}=U^{T}$.