We have that in $\mathbb{R}^3$ the intersection of a sphere and a plane is a circle or a point (or an empty set).
In higher dimentions, i.e. $\mathbb{R}^n$, $n>3$ we take a sphere $\{(x_1, ..., x_n)\mid \sqrt{x_1^2+...+x_n^2}=a\}$ and a plane $\{(x_1, ..., x_n)\mid x_1+...+x_n=b\}$. Now what can they intersection look like?
By analogy, I would guess that the possibilities are a point and a sphere in $\mathbb{R}^n$ (or an empty set), but I don't see how to show that.
To simplify things, you can assume the $n$-sphere is the unit $n$-sphere (meaning $a=1$) and that the plane is perfectly flat say with $x_1=b$. If $b=\pm 1$ then the intersection is a point since $\sqrt{1+x_2^2+...+x_n^2}=1$ implies $x_2=x_3=...=x_n=0$. Similarly, if $-1<b<-1$ then the intersection is a $(n-1)$-sphere with a radius of $\sqrt{1-b^2}$ since $$\sqrt{x_2^2+...+x_n^2}=\sqrt{1-b^2}$$ And obviously, if $|b|>1$ then there's no intersection.