The only surface with at least three distinct lines through each of its points is the plane Wikipedia's article on ruled surfaces includes this remark:

The plane is the only surface which contains at least three distinct lines through each of its points.

I'd like to see a reference or proof for this fact.
I assume that the "surfaces" we consider in this proposition are two-dimensional smooth submanifold of $\Bbb R^3$.
This may also answer this question.
 A: The statement is about smooth surfaces in $\mathbb R^3$ i.e. two-dimensional submanifolds, and the result is that the submanifold is an affine plane. To prove this, let $M$ be the surface and take a point $x\in M$. Then the second fundamental form at $x$ is a symmetric bilinear form $II$ on the tangent space $T_xM$, which is a two-dimensional subspace of $\mathbb R^3$. For a unit vector $v\in T_xM$, it is well known that $II(v,v)$ is the so-called normal curvature. This is given by taking the plane in $T_xM$ spanned by $v$ and the unit normal at $x$, intersecting it with $M$ and taking the curvature of the resulting curve in that plane. In particular, if there is an affine line $\ell=\{x+tv:t\in\mathbb R\}$ which is (locally) contained in $M$. Then $\ell$ coincides with the intersection of the normal plane with $M$, so $II(v,v)=0$. If there are three lines through $x$ contained in $M$, then $II$ vanishes on three one-dimensional subspaces of the two-dimensional space $T_xM$. Linear algebra then implies that $II$ has to vanish at $x$. (Definite forms do not vanish on any one-dimensional subspace, indefinite non-degenerate ones on two and non-zero degerate forms on one such subspace.) If this is true for all $x\in M$ then $II$ vanishes identically and it is well known that this implies that $M$ is a part of a plane. 
