The following terms are defined as in the book Differential Geometry by Do Carmo.
Let $S$ be a regular surface, $p \in S,$ $N$ be a normal vector at $p$ and $v \in T_pS$. Let $P$ be the plane parallel to the plane spanned by $v$ and $N$ and passes through $p$.
Now, I want to show that $P$ intersects with $S$ at a regular curve.
I first tried to work on projections but that does not bring me to the conclusion.
After that, I used the fact that a regular surface is locally a graph. Without loss of generality, I assumed that $S$ is a graph of a function of the form $z=f(x,y)$. The equation of the plane $P$ is $(\vec{x}-p)\bullet(N \times v)=0$, so I put in $\vec{x}=(x,y,f(x,y))$ and get $((x,y,f(x,y))\bullet(u\times v)=0$. I try to use Implicit Function Theorem on this equation but I have no idea to proceed.
Can anyone give me the idea of solving such problem? Thank you.
Introduce new coordinates $(x,y,z)$ in the neighborhood of $p$ such that $p=(0,0,0)$ and $N=(0,0,1)$. The tangent plane $T_pS$ then is the $(x,y)$-plane, and by the implicit function theorem around $p$ the surface $S$ then appears as a graph $z=f(x,y)$, whereby $$f(0,0)=0,\quad f_x(0,0)=f_y(0,0)=0\ .$$ In this setting a typical vector $v\in T_pS$ is given by $v=(\cos\phi,\sin\phi,0)$, and the plane $P:=N\vee v$ is orthogonal to the $(x,y)$-plane. In the neighborhood of $p$ the intersection $\gamma:=S\cap P$ possesses the parametrical representation $$\gamma:\quad t\mapsto {\bf x}(t)=\bigl(t\cos\phi, t\sin\phi,f(t\cos\phi,t\sin\phi)\bigr)\ ,$$ satisfying $${\bf x}'(t)=(\cos\phi,\sin\phi, \ldots)\ne{\bf 0}\ ;$$ hence $\gamma$ is a $C^1$-curve.