Let $u$ be a smooth function defined on the unit sphere $S^2$. Does there exist a plane $P$ passing through the origin such that $P\cap S^2$ contains at least three points $x_1,x_2,x_3$ with $\nabla u(x_i) \cdot n =0$, $i=1,2,3$, where $n$ is a vector normal to the plane $P$?
By taking $P$ to be the plane passing through maximum and minimum points of $u$ we can trivially get two points $x_1,x_2$ with $\nabla u(x_i) \cdot n =0$, $i=1,2$. I wonder if there exist 3 such points.
EDIT: At first, I misunderstood the question and was thinking about planes orthogonal to $\nabla u$, that is great circles on which the restriction of $u$ has at least three critical points. The text below gives an answer to this question, not the one asked by OP.
Here is a sketch of proof that such a plane exists. Let $c \in \mathbb{R}$ be such that the area of the domain $\{x \in \mathbb{S}^2 \mid u(x) \le c\}$ is $2\pi$. Let $M = u^{-1}(c)$ be the level curve of $c$ (may be disconnected). Any great circle tangent to $M$ in a non-inflection point intersects $M$ in at least three points. Thus the restriction of $u$ to this circle takes the same value at least three times, therefore it has at least three critical points.
