Let $$A = \begin{bmatrix} 3 & -6 & 0\\ -6 & 0 & 6 \\ 0 & 6 & -3\end{bmatrix}$$. We show that $$A$$ is orthogonally diagonalizable by finding an orthogonal matrix $$U$$ and a diagonal matrix $$D$$ such that $$A = UDU^\mathsf{T}$$.

The characteristic polynomial of $$A$$ is \begin{eqnarray*} p_A & = & \left| \begin{array}{ccc} 3-\lambda & -6 & 0 \\ -6 & -\lambda & 6 \\ 0 & 6 & -3 -\lambda \end{array}\right| \\ & = & (3-\lambda)\left| \begin{array}{cc} -\lambda & 6\\6 & -3-\lambda\end{array}\right |- (-6)\left| \begin{array}{cc} -6 & 6\\0 & -3-\lambda\end{array}\right | \\ & =& (3-\lambda)(\lambda^2 + 3\lambda - 36) + 6(6\lambda + 18) \\ & =& -\lambda^3 + 81\lambda \\ & =& -\lambda(\lambda - 9)(\lambda + 9) \end{eqnarray*} Hence, the eigenvalues are $$0$$,$$9$$, and $$-9$$.

The RREF of $$A$$ is $$\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{bmatrix}$$. Hence, a basis for $$N(A)$$ is given by the single vector $$v_1 = \begin{bmatrix} 1 \\ \frac{1}{2} \\ 1\end{bmatrix}$$. But $$v_1\cdot v_1 = \frac{9}{4}$$. So we take $$u_1 = \frac{1}{\|v_1\|}v_1 = \frac{2}{3} v_1 =\begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3}\end{bmatrix}$$.

The RREF of $$A-9 I$$ is $$\begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}$$. Hence, a basis for $$N(A-9 I)$$ is given by the single vector $$v_2 = \begin{bmatrix} -2 \\ 2 \\ 1\end{bmatrix}$$. But $$v_2\cdot v_2 = 9$$. So we take $$u_2 = \frac{1}{\|v_2\|}v_2 = \frac{1}{3} v_2 =\begin{bmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3}\end{bmatrix}$$.

The RREF of $$A-(-9) I$$ is $$\begin{bmatrix} 1 & 0 & \frac{1}{2}\\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}$$. Hence, a basis for $$N(A-(-9) I)$$ is given by the single vector $$v_3 = \begin{bmatrix} -\frac{1}{2} \\ -1 \\ 1\end{bmatrix}$$. But $$v_3\cdot v_3 = \frac{9}{4}$$. So we take $$u_3 = \frac{1}{\|v_3\|}v_3 = \frac{2}{3} v_3 =\begin{bmatrix} -\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3}\end{bmatrix}$$.

Hence, setting $$U = \frac{1}{3} \begin{bmatrix} 2 & -2 & -1 \\ 1 & 2 & -2 \\ 2 & 1 & 2 \end{bmatrix}$$ and $$D = \begin{bmatrix} 0 & 0 & 0\\0 & 9 & 0\\ 0& 0& -9 \end{bmatrix}$$, we get $$A = UDU^\mathsf{T}$$.