I think I understood the question:
A plane is tangent to a surface embedding in R^3, at each point of a curve on the surface. Prove the tangent vector to the curve is a principle direction, and the second fundamental form is degenerate( singular, rank 0, or zero determinant) along the curve.
The easiest way to understand the problem is, using the Gauss Weingarten map, N. Normal vectors of surface along the curve is constant, actually they're all identical to the normal vector to the touching plane. So the curve is mapped to a single point by the Gauss map N. So the induced linear map in the tangent space, i.e. the differential dN, maps the tangent vector v of the curve to a 0 vector.
So dN is degenate, with v as a eigenvector, belonging to the eigenvalue 0. And note that dN is nothing but the second fundamental form, we're done.
More reference goes here Shape operator