Q:

Find $f(x,y,z) = 0$ that excludes $q$ and $v$ parameters

I have the following problem: 
$$x = e^q \sin (q) + v$$
$$y = e^q\cos (q) + 2v$$
$$z = e^q + 3v$$
I want to find a function $$f(x, y, z) = 0$$, that is i want to remove the $q$ and $v$ parameters from the equation.
How can I do it ?
Here is my attempt: $$ (x-v)^2 + (y -2v)^2 = e^{2q}$$
$$ (z-3v)^2 = e^{2q}$$
from which I get:
$$x^2 + y^2  - 2vx -4vy = z^2 -6vz + 4v^2$$
Now $q$ is excluded but i don't know how to exclude $v$ too.

A:

Here $$v(x,y,z)=3z-x-2y\pm \sqrt{(3z-x-2y)^2-z^2+x^2+y^2}.$$
and $$q(x,y,z)=\arctan \frac{x-v}{y-2v}+\pi k,$$ so
$$f(x,y,z)=e^{q(x,y,z)}+3v(x,y,z)-z=0.$$

