<s>[INST] Orthogonal change of variables - elimination of cross terms 
Eliminate cross terms of $Q$ and express $Q$ in terms of the new variables.

$$\textit{Q}=2x^2+5y^2+5z^2+4xy-4xz-8yz$$
  $$\begin{pmatrix}
2 &2  &-2 \\ 
2 &5  &-4 \\ 
-2 &-4  &5 
\end{pmatrix}$$
$$-\lambda^3+12\lambda^2-21\lambda +10$$
$$\lambda _1=1, v_1=\begin{pmatrix}
-2 &1  &0 
\end{pmatrix}$$
  $$\lambda _2=1, v_2=\begin{pmatrix}
2 &0  &1 
\end{pmatrix}$$
  $$\lambda _3=10, v_3=\begin{pmatrix}
-1 &-2  &2 
\end{pmatrix}$$

Do I have enough information to make a solution?
  $${\textit{Q}}'={x}'+{y}'+10{z}'$$
  or do I need to normalize and find the orthonormal bases -if so where to start from here?

 [/INST] 
Updated:
Note that $\lambda_{1}=\lambda_{2}$, $\boldsymbol{v}_{1} \perp \boldsymbol{v}_{3}$ and $\boldsymbol{v}_{2} \perp \boldsymbol{v}_{3}$, $\boldsymbol{v}_{1}$ and $\boldsymbol{v}_{2}$ can be resolved into linear combination of $\boldsymbol{e}_{1}$ and $\boldsymbol{e}_{2}$.

Let $\boldsymbol{e}_{1}=\displaystyle \frac{1}{\sqrt{5}}
\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$
and $\boldsymbol{e}_{3}=\displaystyle \frac{1}{3}
\begin{pmatrix} -1 \\ -2 \\ 2 \end{pmatrix}$, then
$$\boldsymbol{e}_{2}=\boldsymbol{e}_{3} \times \boldsymbol{e}_{1}=
\begin{pmatrix} -\frac{2}{3\sqrt{5}} \\ \frac{\sqrt{5}}{3} \\ \frac{4}{3\sqrt{5}}  \end{pmatrix}$$
Therefore
\begin{align*}
  \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} &=
  \begin{pmatrix}
    \frac{2}{\sqrt{5}} & -\frac{2}{3\sqrt{5}} & -\frac{1}{3}  \\
    0 & \frac{\sqrt{5}}{3} & -\frac{2}{3} \\
    \frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{2}{3} \\
  \end{pmatrix}^{-1}
  \begin{pmatrix} x \\ y \\ z \end{pmatrix} \\
  &=
  \begin{pmatrix}
     \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
    -\frac{2}{3\sqrt{5}} & \frac{\sqrt{5}}{3} & \frac{4}{3\sqrt{5}} \\
    -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
  \end{pmatrix}
  \begin{pmatrix} x \\ y \\ z \end{pmatrix}
\end{align*}
$$
\begin{pmatrix}
  \frac{2}{\sqrt{5}} & -\frac{2}{3\sqrt{5}} & -\frac{1}{3}  \\
  0 & \frac{\sqrt{5}}{3} & -\frac{2}{3} \\
  \frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{2}{3} \\
\end{pmatrix}
\begin{pmatrix} 2 & 2 & -2 \\ 2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}
\begin{pmatrix}
  \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
  -\frac{2}{3\sqrt{5}} & \frac{\sqrt{5}}{3} & \frac{4}{3\sqrt{5}} \\
  -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
\end{pmatrix} =
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 10 \end{pmatrix}$$
$$Q = x'^2+y'^2+10z'^2$$
</s>