Is this curve in the plane? (what does this question mean). I've been faced with this question: 

Prove or disprove:
The following curve is contained in the/a plane: (Not sure if a/the) 
$\vec l(t)=(t^2, 2t^2+t,t+1 )$.

Note: The word there is tricky and i couldn't figure out if the meaning was a plane or the plane.

But anyway, I wanted to make sure that I could solve this question correctly on both ways, and here's what I thought of: 
So let's say it was the plane (xy plane) , I thought that the $z$ of the curve must be always $0$ in order for that to happen. so $t+1=0$ and so $t=-1$, but I couldn't reach any conclusion according to this..? 
And if it was just a plane as a random plane: 
I would need that the tangent vector on the curve always perpendicular to the plane's normal. 
So $\vec l'(t)=(2t,4t+1, 1)$. if I take $\vec N=(a,b,c)$. Then $(2t,4t+1,1)\cdot(a,b,c)=2ta+4tb+b+c=0$. 
And I got stuck again not knowing how to find the normal of the plane.
 A: Note that, for each $t\in\Bbb R$,$$2t^2-(2t^2+t)+(t+1)=1.$$Therefore, your curve is a plane curve since, for each $t\in\Bbb R$,$$\vec l(t)\in\{(x,y,z)\in\Bbb R^2\mid2x-y+z=1\}.$$
A: Your second approach is promising. Note that the “curve” here can be seen as the set of all the image points. In this case, the set
$$C = \{\vec{l}(t) : t\in\mathbb{R}\},$$
(I have assumed that the domain of $t$ is $\mathbb{R}$, in other words, that, in this case, $t$ ranges from $-\infty$ to $+\infty$). In any case, and following your second approach, the curve is contained is some plane iff there exists a constant nonzero vector $\vec{N}\in\mathbb{R}^3\setminus \{0\}$ which is normal to the curve (for all $t\in\mathbb{R}$) (this statement is immediate to prove). In this case, the curve would be contained in the plane perpendicular to $\vec{N}$.
Finally, note that, following what you did, you would need to find $(a,b,c)\in\mathbb{R}^3\setminus \{0\}$ such that
$$2ta + (4t + 1)b + c = (2a+4b)t + b + c = 0 \ \ \forall t\in\mathbb{R}$$
Two polynomial functions in $\mathbb{R}$ are equal iff their coefficients are equal. In this case, the previous equation implies:
\begin{align}
\begin{cases}
2a + 4b = 0\\
b + c = 0
\end{cases}
\end{align}
which yields the simple nonzero solution $\vec{N} = (a,b,c) = (2,-1,1)\neq 0$. We have hence found the desired normal vector, and therefore, the curve is contained in the plane perpendicular to $\vec{N}$, namely
$$\pi = \{(x,y,z)\in\mathbb{R}^3 \colon 2x-y+z=0\}$$
