Which is the intersection of the sphere $S^2$ and the plane $x_a(\lambda , \mu)=(a, \lambda , \mu), \ a\in (0,1), \ \lambda , \mu\in \mathbb{R}$ ?
If the plane goes through the sphere the intersection is a circle, right? It is also possible that the intersection is a point. Otherwise there is no intersection.
So we have tocheck the distance of the plane from the center of the sphere, if it is less than the radius, equl or greater, right?
Howcan we do that in this case?
I'm assuming that $$ x_a(\lambda, \mu) = \big\{(a,\lambda,\mu) \in \mathbb R^3: \lambda \in \mathbb R, \mu \in \mathbb R \big\} $$ In other words, $x_a(\lambda, \mu)$ is just a ridiculously complicated way to refer to the plane $x=a$. It's especially goofy since we're given a parametric equation for the plane, and the parameterization has no affect whatsoever on the intersection with the sphere.
Where this plane intersects the sphere $S^2 = \big\{(x,y,z) \in \mathbb R^3: x^2 + y^2 + z^2 = 1 \big\}$, we have $a^2 + y^2 + z^2 = 1$ and so $y^2 + z^2 = 1 - a^2$. So, the intersection is a circle lying on the plane $x=a$, with radius $\sqrt{1-a^2}$. The radius expression $\sqrt{1-a^2}$ makes sense because we're told that $0 < a < 1$.
If $a=1$, then the intersection circle degenerates into a point at $(1,0,0)$.
If $a >1$ or $a<-1$, there is no intersection.