I'm given that the conical surface $f(x,y,z)=3x^2+3y^2+z^2-6xy-2xz+2x-1$ with apex the point $K=(1,1,1)$ and I'm asked to find the equation of the curve (not necessarily a circle) used to make the conical surface?
How do I handle this?
Also if I'm given both the equation of the curve and of the cone how do I find the apex?
Giving the cone $f(x,y,z)=0$ choosing a non-tangent suitable plane
$$ \Pi\to ax+by+cz+d=0 $$
the intersection $f(x,y,z)\circ \Pi(x,y,z)$ gives a conic curve which can be used as supporting curve to create $f(x,y,z)=0$
In the present case we have
$$ f(x,y,z) = 3 x^2 + 3 y^2 + z^2 - 6 x y - 2 x z + 2 x - 1 $$
and choosing a plane as
$$ x+y+z=-15 $$
we have the intersection projection onto the $x\times y$ plane as
$$ 224 + 62 x + 6 x^2 + 30 y - 2 x y + 4 y^2 = 0 = (p-p_0)M(p-p_0)+c_0 $$
with
$$ p=(x,y)\\ p_0 = \left(-\frac{139}{23},-\frac{121}{23}\right)\\ M = \left(\begin{array}{cc}6 & -1\\ -1 & 4\end{array}\right)\\ c_0 = -\frac{972}{23} $$
NOTE
Choosing the plane $z = 0$ the intersection curve is given by
$$ 3x^2+3y^2-6xy+2x-1=0 $$
which is a slanted parabola