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?

 A: 
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*}
$$
\begin{pmatrix}
  \frac{2}{\sqrt{5}} & -\frac{2}{3\sqrt{5}} & -\frac{1}{3}  \\
  0 & \frac{\sqrt{5}}{3} & -\frac{2}{3} \\
  \frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{2}{3} \\
\end{pmatrix}
\begin{pmatrix} 2 & 2 & -2 \\ 2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}
\begin{pmatrix}
  \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
  -\frac{2}{3\sqrt{5}} & \frac{\sqrt{5}}{3} & \frac{4}{3\sqrt{5}} \\
  -\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
\end{pmatrix} =
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 10 \end{pmatrix}$$
$$Q = x'^2+y'^2+10z'^2$$
A: You’ve got the right start, but to diagonalize a quadratic form, you do indeed need to find an orthonormal basis so that you end up with an orthogonal change-of-basis matrix. To see why this is necessary, let $A=B\Lambda B^{-1}$, where $A$ is the matrix of $Q$ and $\Lambda$ is diagonal. We then have $$(B^{-1}\mathbf x)^T\Lambda(B^{-1}\mathbf x)=\mathbf x^T(B^{-1})^T\Lambda B^{-1}\mathbf x$$ so we must have $B^T=B^{-1}$ for this to equal $Q$.
Fortunately, you’re working in $\mathbb R^3$, so you don’t need to go through the Gram-Schmidt process to find an orthonormal basis (though in this case, that’s not much work and gives the same result). For a real symmetric matrix, eigenspaces of different eigenvalues are orthogonal, so we know that $v_1\perp v_3$. Normalize these two vectors and then take their cross product to find a unit vector orthogonal to them both. This vector must also be an eigenvector of $1$, so you have your basis. Specifically, $$\begin{align}u_1&=\left(-\frac2{\sqrt5},\frac1{\sqrt5},0\right)^T\\u_3&=\left(-\frac13,-\frac23,\frac23\right)^T\\u_2&=u_1\times u_3=\left({2\sqrt5\over15},{4\sqrt5\over15},\frac{\sqrt5}3\right)^T\end{align}$$ yielding $$U=\pmatrix{u_1&u_2&u_3}=\pmatrix{-\frac2{\sqrt5}&{2\sqrt5\over15}&-\frac13\\\frac1{\sqrt5}&{4\sqrt5\over15}&-\frac23\\0&\frac{\sqrt5}3&\frac23}.$$ I’ll leave it to you to verify that with $\mathbf x'=U^T\mathbf x$, $x'^2+y'^2+10z'^2=Q$.  
Note, by the way, that once you know that the eigenvalues are $10$, $1$ and $1$, you can save some work by using the fact that the eigenspace of $1$ is the orthogonal complement of the eigenspace of $10$, which allows you to read a basis for it directly from an eigenvector of $10$.
