Let $S$ be the sphere of radius $2$ centered at the origin in $\mathbb{R^3}$. Let $C$ be the circle obtained by intersecting $S$ with the plane $\{x-y+z=3\}$. Parametrize $C$.
If I let $y=x+z-3$ and try to solve it then I run into complications because of the appearance of a cross-term involving two variables. How can I solve this problem? I am having real difficulty finding with this and I would appreciate some help.
I believe the solution assumes that you would know $C$ is a circle. Finding a parametric equation for an unknown shape is probably too difficult. You need to use your intuition to guess the type of the shape, then you can confirm your intuition with algebra later.
So, by intuition, you would have to guess (by drawing or thinking in your head) that $C$ is a circle centered at $\left(1, -1, 1\right)$ and lies on the plane $x - y + z = 3$. Another piece of information needed to define the circle is the radius, which you can guess that it is $\sqrt{2^2 - \sqrt{3}^2} = 1$. (The sphere has radius $2$, and the center of $C$ is $\sqrt{3}$ unit from the origin.)
With my guess, I can parametrize my circle quite easily. Let $u = (1, -1, 1)$, $v = \left(\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}, 0\right)$ and $w = \left(-\frac{1}{\sqrt 6}, \frac{1}{\sqrt 6}, \frac{2}{\sqrt 6}\right)$. (I choose $u$ to be the normal vector of the plane, then $v$ and $w$ to be unit vectors that form an orthogonal basis with $u$. You can pick different $v$ and $w$.) The parametrization of the circle is
$$ p(t) = u + v\cos(t) + w\sin(t) $$ where $t$ is the parameter. In terms of components, I can write \begin{align} x(t) & = 1 + \frac 1{\sqrt 2}\cos(t) - \frac 1{\sqrt 6}\sin(t)\\ y(t) & = -1 + \frac 1{\sqrt 2}\cos(t) + \frac 1{\sqrt 6}\sin(t) \\ z(t) & = 1 + \frac 2{\sqrt 6}\sin(t) \end{align}
Now I am going to verify that my guess is correct. First, I check that $x - y + z = 3$, which is quite easy. Next, I check that $x^2 + y^2 + z^2 = 4$.
\begin{align} x^2 + y^2 + z^2 = p \cdot p & = (u + v \cos(t) + w \sin(t)) \cdot (u + v \cos(t) + w\sin(t)) \\ & = u \cdot u + \cos(t)^2 v \cdot v + \sin(t)^2 w \cdot w \\ & = 3 + \cos(t)^2 + \sin(t)^2 = 4. \end{align}