Find the diametral plane of the surface $$x^2+2y^2-z^2+2xy-2yz-2xz-4x-1=0$$ that is perpendicular to the plane $$x+y+z-3=0$$
I know that the diametral plane passes through the center of the surface (if exists), so I have found the center $C(3,-2,-1)$. So if we denote the plane by $$Ax+By+Cz+D=0$$ then we have $$3A-2B-C+D=0$$ and, also, as it is perpendicular to the given plane $$A+B+C=0$$ Now, I need one more equation, but cannot find it. Any help is appreciated.
$\textbf{Edit:}$ Consider the surface given by the equation $$F(x,y,z)=a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{23}yz+2a_{31}zx+2a_{14}x+2a_{24}y+2a_{34}+a_{44}=0$$ Then, to find the center, one may use the formula $$ \begin{cases} a_{11}x+a_{12}y+a_{13}z+a_{14}=0 \\ a_{21}x+a_{22}y+a_{23}z+a_{24}=0 \\ a_{31}x+a_{32}y+a_{33}z+a_{34}=0 \end{cases} $$ Note that this formula can also be written in the form of partial differentiation by each variable.
In our case, this gives the center $C(3,-2,-1)$