I have a following mathematical problem:
Let's consider a curve: $$\alpha \left(t \right) = \left(\frac{1}{4}t^4 , \frac{1}{3}t^3 , \frac{1}{2}t^2 \right) , t \in R$$ Find an equation of tangential axis of the curve $\alpha$ which is parallel to a plane $x + 3y + 2z = 0$.
Apparently, I don't know what 'tangential axis of a curve' means. During my previous exercises, I have encountered 'a normal axis of a curve' - a straight line described by a normal vector. In this case I just had to calculate $\alpha'' \left( t \right)$ and put it into an equation of a straight line $t\alpha'' \left( t_{0} \right) + \alpha \left( t_{0} \right)$. However, in this case it doesn't seem to make any sense because if a calculate tangent $\alpha' \left( t_{0} \right)$ and insert it into an equation of a line then I don't see any way how I can make it parallel to a plane (because vector $\alpha' \left( t_{0} \right)$ already describes if it is parallel to a plane or not).
Is my intuition about a tangential axis to a curve correct?