For the function $$ \vec x(t) = \begin{pmatrix} 2t + 3 \\ 2 - t \\ t^3 - 2t^2 + t \\ \end{pmatrix} t ≥ 0$$ Find all points $\vec x(t_0)$ for which the tangent line passes through the point (1, 3, 0).
I understand how I would go about solving the problem if I had a normal polynomial function or only x and y values. However, with x, y, and z values I am not sure how to approach the question.
I know that the derivative of the function is
$$ \vec x'(t) = \begin{pmatrix} 2 \\ -1 \\ 3t^2 - 4t + 1 \\ \end{pmatrix} $$
Would I then have to find the equation of the tangent line and use that with the given point to find the other points?
Hint:
Just write the vectors $\vec x(t)-(1,3,0)$ and $\vec x'(t)$ are collinear: $$\frac{2t+3-1}2=\frac{-t-1}{-1}=\frac{t^3-2t^2+t}{3t^2-4t+1}\iff t+1=\frac{t^3-2t^2+t}{3t^2-4t+1}=\frac{t(t-1)^2}{(3t-1)(t-1)}=\dotsm$$