How to pick a solution when there are infinitely many So I am studying this particular set of equations $$x+y+z=2$$
$$x^2+y^2+z^2+2xy=1$$
which can be combined to make
$$2x^2+2y^2+2z^2+4xy-x-y-z=0$$
Putting this into Wolfram gives me an infinite elliptic cylinder

This was to be expected, of course, since the system is underdetermined. However, I now want to pick a point $(x,y,z)$ from off the surface. My questions are

*

*In a situation like this, is there a notion of a best point under some criterion?

*Further, how would one even go about picking a point? I'm worried that just choosing numbers randomly might land me in a situation akin to $x^2+y^2=0$, where if I chose, for example, $y=1$ by chance, then $x$ would be complex, which is not desirable in this case.

*Finally, could the proposed method be extended to surfaces in $\mathcal{R}^n$?

I have already tried combing through the internet and this site, but I could only find a solution if the object was a sphere, not an elliptic cylinder. Any help or resources would be appreciated. Thank you!
 A: This is a comment, but it's too long to fit in a comment box.
The first thing you should know is that, while your third equation is derived from the previous two equations, and thus contains all the solutions to your system, it is not equivalent to your previous two equations (i.e. it contains a lot more solutions).
Instead, consider that
$$x^2 + y^2 + z^2 + 2xy = 1 \iff (x + y)^2 + z^2 = 1.$$
Let $w = x + y$. Then our system becomes
$$\begin{cases} w + z = 2 \\ w^2 + z^2 = 1.\end{cases}$$
Unfortunately, this system has no solution. There are multiple ways to show this, but probably the most elementary method is to substitute $z = 2 - w$ into the second equation:
$$w^2 + (2 - w)^2 = 1 \iff 2w^2 - 2w + 1 = 0,$$
a quadratic with negative discriminant. Thus, no real solutions exist.
So, your first two questions are moot.
A: I guess, more research is needed.
$$
\begin{align*}
x + y + z &= 2 \tag{1}\\
x^2 + y^2 + z^2 + 2xy &= 1 \tag{2}
\end{align*}
$$
Consider eq.(1)
$$
(x + y + z)^2 = 2^2
$$
$$
x^2 + y^2 + z^2 + 2xy + 2yz + 2 zx = 4 \tag{3}
$$
Now subtract the eq.(2)
$$
2yz + 2 zx = 3 \tag{4}
$$
Subtract twice eq.(4) from eq.(3)
$$
x^2 + y^2 + z^2 + 2xy - 2yz - 2 zx = -2
$$
$$
(-x - y + z)^2 = -2.
$$
The latter means there is no real solution to the system of equations (1) and (2).
