Vertical cross-section of $S : r(u,v) = (u+v,uv,u^2v)$. Exercise :

Consider the surface $S : r(u,v) = (u+v,uv,u^2v), \; (u,v) \in \mathbb R^2$. Express the vertical cross section $c$ of the surface at the point $(2,1,1)$ with direction $(2,1)$ and furthermore calculate the vertical curvature of $c$ at the point $(2,1,1)$.

Question :
I would like to kindly ask how is a vertical cross section defined and most importantly calculated, in differential geometry terms, in such a case. 
I assume that since we are interested in a vertical cross section, it should be a cross section with a plane that the unitary vertical vector of $S$, $\mathbf{N} = \frac{r_u \times r_v}{\|r_u \times r_v\|}$ belongs to it. 
I have calculated the unitary vertical vector at the point given $(2,1,1)$, which is achieved for $(u,v) = (1,1)$ as :
$$N(1,1) = \frac{\sqrt{2}}{2} (-1,2,0)$$
From a note I found, the curves that come up as sections of the surface with planes from the point $p$ which are parallel to $\mathbf{N}$, are named vertical cross sections of $S$ at $p$. 
I assume this is almost what we are interested in. But, where does the direction given come in play ? 
All in all, I can't seem how to proceed with a method to calculate the vertical cross section $c$ as asked. As I am preparing for some exams, I would very much appreciate anyone who could give an insight !
 A: For anyone that may be interested in this post in the future, I worked around the following solution :

The vertical cross section curve that we're being asked to find, is the event of a cross section with a plane $(\pi)$ that consists of the unitary vertical vector of $S : \mathbf{N} = \frac{r_u \times r_v}{\|r_u \times r_v\|}$ as well as the point given $(2,1,1)$ and an "essence" of the direction $(2,1)$. I will find vectors $\mathbf{a}$ and $\mathbf{b}$ that will help us express the plane $(\pi)$. To start off, note that $(2,1,1) = r(1,1)$.
$$\mathbf{a} \parallel \mathbf{N} \implies \mathbf{a} = r_u(1,1)\times r_v(1,1) = (-1,1,0)$$
$$\mathbf{b}\; \parallel \;\text{direction} \implies \mathbf{b} = 2r_u(1,1) + r_v(1,1) = (3,3,5)$$
Thus, an expression for $(\pi)$ would be :
$$(\pi) : f(s,t) = (2,1,1) + s\mathbf{a} + t\mathbf{b} = (2-s+3t,1+s+3t, 1+5t)$$
Thus, the curve derived by the vertical cross section described would be :
$$c : f(s,t) = r(u,v)  \implies \begin{cases} u+v &= 2-s+3t \\ uv &= 1+ s + 3t \\ u^2v &= 1+5t\end{cases}$$
