$x^2+y^2+z^2=4$ and $x+y+z=3$
First I tried to parameratize:
$t=x^2 \to x=\sqrt{t}$
$t=y^2 \to y=\sqrt{t}$
Then substituting those parameters into the plane to get:
$z=3-\sqrt{2t}$
These three parameters satisfy the equation for the plane yielding 3=3, but plugging these into the sphere should yield the point(s) of intersection, I believe.
Best Answer
the plane and the sphere intersect to form a circle.
the parameterized equations of the circle are going to be something along the lines of
$\mathbf x(\theta) = \mathbf u \cos\theta + \mathbf v \sin\theta + \mathbf c$
Where $\mathbf c$ is the center of the circle (follow the vector normal to the plane from the center of the sphere until it intersects the plane to find the center of the circle), and $\mathbf u, \mathbf v$ are orthogonal vectors in the plane.
$\mathbf c = (1,1,1)$
What is the radius of the circle?
$\|\mathbf c\| = \sqrt 3\\ 3 + r^2 = 4\\ r = 1$
$\mathbf u = (\frac 1{\sqrt 2},-\frac1{\sqrt 2},0)$ would be a unit vector in the plane, and $\mathbf v = (\frac 1{\sqrt 6}, \frac 1{\sqrt 6},-\frac 2{\sqrt 6})$ would be a unit vector in the plane that is orthogonal to $\mathbf u$.