<p>Okay.
I am not exactly sure what the simplest way of doing this is.
But just on the face of it (and I can't be bothered to actually check this), it seems to make sense to do the following:
By multiplying the first equation by 2 and the subtracting by the second equation, you can express x as a quadratic function of y. Let this function be f(y).
By putting this back into either of the initial equations, you can express z^2 as a quartic function of y. Let this function be g(y).
Any plane can be expressed as z = ax + by + c as someone pointed out before in this thread.
So suppose there exists such a plane which completely contains the curve.
Since the curve is continuous, there exist infinitely many sets of values (x, y, z) which satisfies both the plane equation and f(y) and g(y).
So this must mean that there are infinitely many solutions to the polynomial equation g(y) = (a*f(y) + by + c)^2 (by squaring the plane equation and then replacing z^2 by g(y) and x by f(y).)
So the L.H.S. must be identical to the R.H.S.
So we can equate coefficients of the polynomial and solve for a, b and c.
If this is possible, then there indeed exists such a plane.
If it is impossible to do this, (since this is a quartic equation, there will be five constants/coefficients to determine and only 3 variables, so there may not exist a solution set) then there doesn't exist such a plane.</p>
<p>I have a bad feeling that either I made a mistake somewhere, or that there might not exist a solution, but I don't know since I haven't actually checked it.
I must admit, I have left maths for quite some time since the IMO in July, and haven't done any non-IMO maths (like calculus or vectors) for a long time like a year or so.
Still, I am embarrassed that I can't give a really easy answer.
I should really get back into it.
It is weird that my 'solution' is more like a linear algebra/vector geometry solution more than a multivariate calculus one. (Another reason why I think there must be a much easier way which I for some humiliating reason can't see)
Whatever.
I mean, I am ashamed, but I really need to go right now and I'll maybe come back when I have time and have another look (if I remember).</p>