IAS math problem 2017-paper 1 4a Reduce the following equation into standard form and hence determine the nature of coincoid $x^2+y^2+z^2-zx-xy-yz-3x-6y-9z+21=0$
I have reduced the equation into $(x-3/2)^2 +(y-3)^2 + (z-9/2)^2 -zx -xy-yz = 21/2$ from here on I can sense to reduce it to any of the standard 3D form that I am aware of(like ellipse, ellipsoid, hyperboloid, etc.,). Please help me out in the same.
 A: I got that our equation it's
$$(x-y+7)^2+(y-z+1)^2+(z-x-8)^2=36(x+2).$$
I think from here we can  get $x=-2$, but I don't see this.
A: just to find the nature of the surface does not require an orthonormal coordinate change. Take
$$ u = x+y+z, \; \; \; v = -x + y, \; \; \; w = -x - y + 2z \; \; .  $$
$$ 6u = \frac{3}{4} (v-1)^2 + \frac{1}{4} (w-3)^2 + 18 $$
which is a paraboloid. The $6u$ stuff does not show this, but it is a paraboloid of revolution. One could correct $u,v,w,$ in particular dividing by $\sqrt 3, \sqrt 2, \sqrt 6,$ to get a strictly rotated version. The axis of symmetry (actually of revolution, as I said) is parallel to $(1,1,1)$ in the original $(x,y,z)$ coordinates, and the vertex is at $(0,1,2).$
Having done that much, we get an orthonormal change taking $u = p \sqrt 3,$ $v = q \sqrt 2,$ $w = r \sqrt 6.$ 
$$ 6 p \sqrt 3 = \frac{3}{4} (q \sqrt 2-1)^2 + \frac{1}{4} (r \sqrt 6-3)^2 + 18, $$
$$ 6 p \sqrt 3 = \frac{3}{2} (q -\frac{1}{\sqrt 2})^2 + \frac{3}{2} (r  - \frac{3}{\sqrt 6})^2 + 18, $$
$$ 4 p \sqrt 3 =(q -\frac{1}{\sqrt 2})^2 +  (r  - \frac{3}{\sqrt 6})^2 + 12 \; \; , $$
$$  p  = \frac{1}{4 \sqrt 3} \left(\left(q -\frac{1}{\sqrt 2}\right)^2 +  \left(r  - \frac{3}{\sqrt 6}\right)^2 + 12 \right) \; \; . $$
