Given the curve $\displaystyle C$ on $\displaystyle \mathbb R^3$ parametrized by $\displaystyle r(t).$ Find the normal vector on $\displaystyle r(1)=(3,1,3)$ if the osculator plane on that point is $\displaystyle 3y-x=0$ and $\displaystyle r(1)\cdot N(1)>0$ with $\displaystyle r'(1)=(6,2,2).$ (I don't know if "osculator" makes sense here.)

I hope you guys can help, I have an exam on march and I'm scared.

I don't remember how to solve this, since I lost my calculus notebook.

Thanks!