$\vec\alpha : I \rightarrow S \subset \mathbb R^3$ is a curve on a regular surface S
and the ruled surface generated {$\vec\alpha(u), \vec N(u)$}, $\vec N(u)$ is the normal to the surface $S$ at $\vec\alpha$
I need to show that, $\vec\alpha(I) \subset \mathbb S$ is a line of curvature in $S$ $\iff$ this ruled surface is developable.
My Attempt:
$\vec x(u,v) = \vec\alpha(u) + \vec vN(u)$
and
$(\vec N,\vec N',\vec\alpha') = 0$
If $\vec\alpha$ is a line of curvature of $S$ then,
$\vec N'(u)=d\vec N_p(\vec \alpha'(u))=\lambda(u)+\vec \alpha'(u)$
Thanks...
Your ruled surface is parametrized by $r(u,v)=\alpha(u)+vN(\alpha(u))$, so its tangent planes at points $(u,v)$ are spanned by
$$\frac{\partial r}{\partial u}(u,v)=\alpha'(u)+vDN_{\alpha(u)}(\alpha'(u))\text{ and }\frac{\partial r}{\partial v}(u,v)=N(\alpha(u)).$$
The tangent space at $(u,0)$ is spanned by $\alpha'(u)$ and $N(\alpha(u))$. Thus, the surface will be developable if and only if the vector $DN_{\alpha(u)}(\alpha'(u))$ is a linear combination of $\alpha'(u)$ and $N(\alpha(u))$. Since
$$DN_{\alpha(u)}(\alpha'(u))=\underbrace{[DN_{\alpha(u)}(\alpha'(u))]^\top}_{\in T_{(u,v)}S}+\underbrace{[DN_{\alpha(u)}(\alpha'(u))]^\perp}_{\in\,\mathrm{span}(N(\alpha(u)))},$$
this is equivalent to $[DN_{\alpha(u)}(\alpha'(u))]^\top=S_{u}(\alpha'(u))$ being a multiple of $\alpha'(u)$, i.e. $\alpha'$ is made of eigenvectors of the shape operator, or again $\alpha$ is a line of curvature.