<s>[INST] Given 3 points. What's the normal to the plane that contains these 3 points? 
Given are these points $$v_1=\begin{pmatrix} 1\\  2\\  3
\end{pmatrix}, v_2=\begin{pmatrix} 2\\  3\\  3 \end{pmatrix},
v_3=\begin{pmatrix} 2\\  4\\  4 \end{pmatrix}$$
Determine the normal to the plane which contains these 3 points.

My problem with this is I'm not sure what is really meant with "normal". Is it just the normal-vector or I will also need some parameter multiplied with the normal-vector?
So far I've done that but I'm not sure if that is all what is asked?
First calculate both direction vectors:
$$\begin{pmatrix}
2\\ 
3\\ 
3
\end{pmatrix}-
\begin{pmatrix}
1\\ 
2\\ 
3
\end{pmatrix}=
\begin{pmatrix}
1\\ 
1\\ 
0
\end{pmatrix} \text{ and }
\begin{pmatrix}
2\\ 
4\\ 
4
\end{pmatrix}-
\begin{pmatrix}
1\\ 
2\\ 
3
\end{pmatrix}=
\begin{pmatrix}
1\\ 
2\\ 
1
\end{pmatrix}$$
Now cross-product of these two vectors gives the normal vector:
$$\begin{pmatrix}
1\\ 
1\\ 
0
\end{pmatrix}
\times
\begin{pmatrix}
1\\ 
2\\ 
1
\end{pmatrix}=
\begin{pmatrix}
1\\ 
-1\\ 
1
\end{pmatrix}$$
But now is this really an answer to the question? Or I need some parametric form of it? Please help :/
 [/INST] “The” normal to a plane typically really does refer to any vector that is normal to the plane. In other words, your suggested answer (or any non-zero scalar multiple of it) is perfectly fine.
No need to express the normal generally as $$k\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}\quad(k\in\mathbb R\setminus\{0\}).$$
</s>