How to show a curve lies in a plane given 3 constant vectors

Question

How does one show that given 3 non-parallel vectors in $\mathbb{R}^3$, a curve with parameterisation: $$\textbf{s}(t)=(x_1t^2+y_1t+c_1,x_2t^2+y_2t+c_2,x_3t^2+y_3t+c_3)$$ lies in a plane?
Additionally, how do we find the equation of the plane?

I know that any two non-parallel vectors lie on a plane but have no idea when it comes to 3...

Answer 1

Note that $ s(t) = \begin{pmatrix} x_1 & y_1 & c_1\\ x_2 & y_2 & c_2\\ x_3 & y_3 & c_3 \end{pmatrix} \begin{pmatrix} t^{2} \\ t\\ 1 \end{pmatrix}$, and the curve $ \begin{pmatrix} t^{2} \\ t\\ 1 \ \end{pmatrix} $ lies in a plane, and the image of a plane under a linear transformation is a plane.

Answer 2

$s(t)$ is a plane curve iff $\det(s'(t),s''(t),s'''(t))=0$.