<s>[INST] How to solve this quadratic form equation? 
Let $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ be a quadratic form and 
$A = \begin{bmatrix}
       7 & -5 & 4     \\
      -5 & 7  & 4 \\
       4 & 4 & -2
     \end{bmatrix}$ its matrix. Such that $Q=X^TAX$ for $X=\begin{bmatrix} x \\y \\ z       \end{bmatrix}$.
Find an $X$ such that $X^TAX=72$.

I found that the eigenvalues of $A$ are $-6,6\text{ and }12$, so the matrix is not positive definite neither negative definite.
How can I solve this kind of problem? Can you give me a hint? Thanks
 [/INST] I found the follow technic for this kind of problem, that is very similar to the Mark Bennet's answer.
If $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ then $A$ is the associated matrix.
Let $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \neq0 \text{ and } \lambda \in \mathbb{R}$. If $AX=\lambda X$, then $\lambda$ is an eigenvalue of $A$ and $X$ is an eigenvector of $A$, associated to $\lambda$.
So, $Q(x,y,z)=X^TAX= \lambda X^T X= \lambda (x^2+y^2+z^2)$. One have that $(x^2+y^2+z^2) >0$, so if $\lambda (x^2+y^2+z^2)=72$, then $\lambda>0$.
Between the three eigenvalues of $A$, only $6 \text{ and } 12$ can be applied.
Now by finding the vector space span for the eigenvectors for $6 \text{ and } 12$, one got:
$E_{6}=\langle(1,1,1) \rangle$
$E_{12}=\langle(-1,1,0) \rangle$ 
About the $X$ that is asked, or $X \in E_{6}$ or $X \in E_{12}$. I found that $X \in E_{12}$:
Let $\begin{bmatrix} -a \\ a \\ 0 \end{bmatrix}$ be a generic vector of $E_{12}$. So , $12 \left((-a)^2+a^2+0^2 \right)=72$. Then 
$$12 \left(2a^2 \right)=72$$
$$ 24a^2 =72$$
$$a^2=3$$
$$a=\sqrt3$$
Finaly, a $X$ that satisfy what is asked is $X=\begin{bmatrix} -\sqrt3 \\ \sqrt3 \\ 0 \end{bmatrix}$. You can confirm, by computing $Q(x,y,z)$.
$$Q(-\sqrt3,\sqrt3,0)=7 \cdot (-\sqrt3)^2+7\cdot \sqrt3^2 -10 \cdot (-\sqrt3) \cdot \sqrt3$$
$$Q(-\sqrt3,\sqrt3,0)=7 \cdot 3+7\cdot 3+10 \cdot 3$$
$$Q(-\sqrt3,\sqrt3,0)=21+21+30=72$$
</s>