where $I\subset\mathbb R$ is an interval. Show that the image of $c$ is contained in a straight line.
we defined the curvature with;
$\kappa_c(t)= \frac{1}{||\dot{c}(t)||^3}det(\dot{c}(t),{\ddot{c}(t)})$
Since it's regularly parametrized, $\dot{c}(t)≠0$, therefore $det$ must be zero,
Hence $\dot{c}\ and\ \ddot{c} $are linearly dependent $\Rightarrow\lambda\dot{c}=\ddot{c}$???
this has nothing to do with a line.
The formula you want to use is given for a plane curve only (the determinant of two vectors in $\mathbb R^3$ is not well defined).
If $c:I\rightarrow \mathbb R^3$ is a curve, then its curvature is given by $$\kappa_c(t)=\dfrac{\|\dot{c}(t)\times \ddot{c}(t) \|}{\|\dot{c}(t)\|^3}$$
Actually, you can check that the curvature of a curve is independent of the choice of the parametrization. It is a geometric invariant of a curve, so you can choose the arc-length parametrization when the curve is regular.
Assume now that $c$ is parametrized by the arc-length i.e. $\|\dot{c}(t)\|^2=\langle \dot c(t),\dot c(t)\rangle = 1$. Then you can differentiate this last expression to get : $$\langle \ddot{c}(t), \dot c(t)\rangle =0.$$
This means that this two vectors are perpendicular. But, $$\kappa_c(t)=\|\dot{c}(t)\times \ddot{c}(t) \|=0$$ so $\ddot c(t)=0$ for all $t\in I$ since $\dot c(t)\neq 0$.
Hence, $c(t)=c(0)+tc'(0)$ on $I$ and the curve is contained in a line.