The twisted cubic curve $C \subset \mathbb{P}^3$ is described as image of the Veronese map $v: \mathbb{P}^1 \to \mathbb{P}^3$
$$v: [X_0: X_1] \mapsto [X_0^3 : X_0^2X_1 : X_0 X_1^2 : X_1^3] = [Z_0: Z_1: Z_2: Z_3]$$
Alternatively $C= v(\mathbb{P}^1)$ may be also described as zero locus of the polynomials
$$F_0(Z) =Z_0 Z_2-Z_1^2$$
$$F_1(Z) =Z_0 Z_3-Z_1 Z_2$$
$$F_2(Z) =Z_1Z_3-Z_2^2$$
What one can show is that the intersection of $V(F_i)$ and $V(F_j)$ with $i \neq j$ is the union of a line and $C$.
Exercise 1.12.(from Joe Harris' Algebraic Geometry: A First Course. p 10)
Show that any finite set of points on a twisted cubic curve are
in general position, i.e., any four of them span $\mathbb{P}^3$.
One simple way to show it is to observe that every hyperplane in $\mathbb{P}^3$ is given as zero locus of linear polynomial $H= \sum_{i=0}^4 a_i Z_i$ and it suffice to show that if we insert $X_0^3, X_0^2X_1, X_0 X_1^2, X_1^3$ in the $Z_i$ we obtain a homogeneous polynomial of degree $3$ in $X_0, X_1$ which cannot have more than $3$ zeroes in $\mathbb{P}^1$. So no intersection of a hyperplane in $\mathbb{P}^3$ with $C$ hase more than $3$ points. That works.
I would like to know if it also possible to prove 1.12 directly by playing with intersections of hyperplanes $H \subset \mathbb{P}^3$ with the $V(F_0, F_1, F_2)$. That is can it be directly proved that if we intersect the set of solutions of $F_0, F_1, F_2$ with any hyperplane $H=\sum_{i=0}^4 a_i Z_i$ the resulting finite set of solutions consists at most of $3$ points?