Surface described by the equation $-3y^2 - 4xy + 2xz + 4yz - 2x - 2z + 1 = 0$ Given the equation : $-3y^2 - 4xy + 2xz + 4yz - 2x - 2z + 1 = 0$. Check if the surface described by that equation has a center of symmetry and then by making the correct coordinate system change, find the type of the surface that this equation describes.
Attempt : For the symmetry part, we'll just plug in $x\to -x$, $y\to -y$, $z\to -z$ and we will conclude. My question is for the second part of the exercise. I've found some more like these, but ALL of them were described by a quadratic form, which I could write it's Matrix down and compute Eigenvalues-Eigenvectors and then Gram-Schmidt the eigenvectors to find an canonical form and see what surface it was. But in this particular example, I'd do the following :
$-3y^2 - 4xy + 2xz + 4yz  = 2x + 2z - 1$
Now, we have that the equation $f(x,y,z) = -3y^2 - 4xy + 2xz + 4yz$ is a quadratic form that we can compute its Matrix - eigenvalues - eigenvectors and that we can find what surface it describes. Then, this particular surface will be "cut" by the plane $2x+2z - 1 =0$ and thus we can find the surface that the starting equation describes. Is this a correct approach ? If not, please give me some thorough help to understand how to work on these.
 A: Your idea is good but note that the equation 
$$ -3y^2 - 4xy + 2xz + 4yz  = 2x + 2z - 1 $$
does not define the intersection of the surface described by the quadratic form with the plane. Such an intersection would have been described by two equations and not a single equation:
$$ -3y^2 - 4xy + 2xz + 4yz = a, \,\,\, 2x + 2z - 1 = b. $$
What works better is to try and write your equation as $(\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = c$. If $\mathbf{v} = \mathbf{0}$, you get the level set  $q_A(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = c$ of the quadratic form associated to $A$ which you can analyze by your method. If $\mathbf{v} \neq 0$, the picture is the same with the "center" translated to $\mathbf{v}$ which is what I guess you call "the center of symmetry". Note that we have
$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = \mathbf{x}^T A \mathbf{x} - \mathbf{v}^T A \mathbf{x} - \mathbf{x}^T A \mathbf{v} + \mathbf{v}^T \mathbf{A} \mathbf{v} = \mathbf{x}^T A \mathbf{x} - 2 \mathbf{v}^T A \mathbf{x} + \mathbf{v}^T \mathbf{A} \mathbf{v} $$
so the expression splits into a quadratic part in $\mathbf{x}$, a linear part and a constant part. In your case, quadratic part corresponds to: 
$$ A = \begin{pmatrix} 0 & -2 & 1 \\ -2 & -3 & 2 \\ 1 & 2 & 0 \end{pmatrix}.$$
The linear part corresponds to
$$ -2 \mathbf{v}^T A \mathbf{x} = -2(v_1,v_2,v_3)A \begin{pmatrix} x \\ y \\ z \end{pmatrix} =  -2 (v_1, v_2, v_3) \begin{pmatrix} -2y + z \\ -2x - 3y + 2z \\ x + 2y \end{pmatrix} = \\
 -2v_1(-2y + z) -2v_2(-2x - 3y + 2z) -2v_3(x + 2y) = \\
x(4v_2 - 2v_3) + y(4v_1 + 6v_2 - 4v_3) -z(2v_1 +4v_2) \stackrel{\mbox{?}}{=}  -2x - 2z $$
and so 
$$ 4v_2 - 2v_3 = -2, \\
4v_1 + 6v_2 - 4v_3 = 0,\\
 2v_1 + 4v_2 = 2 $$
which implies that
$$ \mathbf{v} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \,\,\, \mathbf{v}^T A \mathbf{v} = 2. $$
Thus, we get
$$ (\mathbf{x} - \mathbf{v})^T A (\mathbf{x} - \mathbf{v}) = q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2. $$
To get your equation, we see that we need to choose $c = 1$ and then
$$ q_A(\mathbf{x} - \mathbf{v}) = -3y^2 -4xy + 2xz + 4yz -2x - 2z + 2 = 1 \iff \\
-3y^2 -4xy + 2xz + 4yz -2x - 2z + 1 = 0. $$
Since $A$ has eigenvalues $-5,1,1$, the surface is a one-sheeted circular hyperboloid (see here).
