For the function $$ \vec x(t) = \begin{pmatrix} 2t + 3 \\ 2 - t \\ t^3 - 2t^2 + t \\ \end{pmatrix} t ≥ 0$$ Are there 2 points $\vec x(t_1)$, $\vec x(t_2)$, such that the function’s tangent vectors at these points are parallel to each other? Find such points, or show that none exist.
I know that the derivative of the function is
$$ \vec x'(t) = \begin{pmatrix} 2 \\ -1 \\ 3t^2 - 4t + 1 \\ \end{pmatrix} $$
and that in order for the tangent vectors to be parallel to each other the functions will equal the same value. However, I am not sure how to go about finding the values.
Notice that $x'(t)=(2,-1,3t^2-4t+1)$ and $g(t)=3t^2-4t+1$ is not injective for $t\geq0$. In particular the equation $g(t)=0$ has two positive distinct solutions.