$\mathcal{C}$ is the zero locus of polynomials which vanish at $p_1,\dots,p_7$ I have to prove that the twisted cubic $\mathcal{C}$ is the common zero locus of quadratic polynomials which vanish in $\{p_1,\dots,p_7\}$, where $p_i \in \mathcal{C}, \forall i=1,\dots,7$.
I saw this answer Why 7 points on a twisted cubic is enough to fix a quadratic?, but I can't understand the last part, where it says " $Q_1 \cap Q_2 \cap Q_3 = \mathcal{C}$, because $Q_3 \cap L \subset \mathcal{C}$ ".
How can I prove that $Q_3 \cap L \subset \mathcal{C}$? And how can I conclude from this that $\mathcal{C} \supset $ {common locus of quadratic polynomials which vanish in $\{p_1,\dots,p_7\}$}?
Can anyone help me?
Thank you in advance
 A: $\newcommand{\PSP}{\mathbb{P}}$
$\newcommand{\Ohol}{\mathcal{O}}$
$\newcommand{\ideal}[1]{\mathfrak{#1}}$
All projective spaces and varieties shall be over an (algebraically closed) field $k$.
Let $\nu:\PSP^1 \to \PSP^3$ be the 3-tuple embedding (Veronese embedding of degree 3) and
$C = \nu(\PSP^1)$ be its image, the twisted cubic curve in $\PSP^3$.
Let $\Gamma = H^0(\PSP^3, \Ohol_{\PSP^3}(2))$ be the 10-dimensional vector-space
of quadrics in $\PSP^3$.
We first prove, that seven pairwise 
different points $P_1,\ldots,P_7 \in C$ give seven independent restrictions
on the space $\Gamma$ leaving a kernel $\Gamma_3 \subseteq \Gamma$ of vector-space-dimension
3:
Let $P_i = \nu(S_i)$ for seven different points $S_i \in \PSP^1$. Then $P_1,\ldots,P_s$
give independent conditions on $\Gamma$ iff $\phi(P_i) = \phi(\nu(S_i))$ correspond to
linear independent vectors in (the affine cone associated to) $\PSP^9$, where $\phi:\PSP^3 \to \PSP^9$ is the 2-tuple embedding
(Veronese embedding of degree 2).
If all $S_i = (1: t_i)$ lie in one affine chart, which is always possible, we simply note,
that the coordinates $(x_0:\ldots:x_9)$ of $\PSP^9$ contain the elements $1,t_i,t_i^2,\ldots,t_i^6$
for $T_i = \phi(\nu(S_i))$. By the non-vanishing of the Vandermonde-matrix-determinant we conclude,
that for $t_1,\ldots,t_7$, all pairwise different, the points $T_1,\ldots,T_7$ impose seven
independent vanishing conditions for linear forms in $\PSP^9$ and therefore $P_1,\ldots,P_7$
seven independent vanishing conditions on the space $\Gamma$ of quadrics on $\PSP^3$.
So we get our space $\Gamma_3$ above of quadrics vanishing on seven points of $C$. By Bezout's theorem
a $Q \in \Gamma$ intersects either $C$ in 6 points or contains it in full. So all quadrics
$Q \in \Gamma_3$ contain $C$.
Two linear independent $Q_1, Q_2 \in \Gamma_3$ intersect in $C \cup L$ with $L$ a line in $\PSP^3$,
because $\deg(Q_1 \cap Q_2) = 2 \cdot 2 = \deg C + \deg L$. Both $C$ and $L$ occur with multiplicity $1$.
This follows from Bezout's theorem (in the form stated for example in Hartshorne's Chapter I).
Intersecting $Q_1 \cap Q_2$ with a third, linear independent, $Q_3 \in \Gamma_3$ gives a priori
$C$ and a part of $L$. If this part is the whole of $L$, then $Q_3$ vanishes on $C \cup L$, which 
has as ideal the intersection $\ideal{p}_C \cap \ideal{p}_L = (Q_1,Q_2)$ (the equality follows because $(Q_1,Q_2)$ has no embedded primary components). So we get
$Q_3 \in (Q_1,Q_2)$ which (by homogeneity) leads to $Q_3 \in k Q_1 + k Q_2$, so $Q_3$ would be
linear dependent on $Q_1,Q_2$ contrary to the assumption.
So we have either $Q_1 \cap Q_2 \cap Q_3 = C$ or it might (seemingly) happen, that $Q_1 \cap Q_2 \cap Q_3 =
(C \cup L) \cap Q_3 = C \cup P$ with a point $P \in L$.
The case, that $Q_3$ intersects $L$ in two points $P'$ and $P''$ not in $C$ would lead to $Q_1 \cap Q_2 \cap Q_3 = C \cup L$. This is because then $Q_1 \cap Q_3 = C \cup L(13)$ with $L(13)$ containing $P'$ and $P''$ and therefore identical to $L$. So we had $L(13) = L(23) = L$ and indeed $Q_1 \cap Q_2 \cap Q_3 = C \cup L$.
Now consider the map $\theta:\PSP^3-C-W \to \PSP^2$ given by $(Q_1:Q_2:Q_3)$, where $C \cup W$ is
the common vanishing set of $Q_1,Q_2,Q_3$ which is, by the above, either $C$ or $C \cup P$.
The fiber $\theta^{-1}(p)$ for a point $p \in \PSP^2$ is the intersection of two linear
independent quadrics from $\Gamma_3$ and therefore of the form $C \cup L$ as seen above.
So we have a map $\PSP^2 \to G(1,\PSP^3)$ with $p \mapsto L(p)$ where $G(1,\PSP^3)$ is the space of lines in $\PSP^3$.
Let us assume, that not all lines $L(p)$ contain a single point $P_0 \in \PSP^3$. Then we have
three points $p_1,p_2,p_3 \in \PSP^2$ such that $\theta^{-1}(p_1) \cap \theta^{-1}(p_2)$ meet
at most in $P_{01}$ and $\theta^{-1}(p_1) \cap \theta^{-1}(p_3)$ meet at most in $P_{02} \neq P_{01}$.
So $\theta^{-1}(p_1) \cap \theta^{-1}(p_2) \cap \theta^{-1}(p_3)$ are empty. The intersection
$\bigcap_{j=1}^3 \theta^{-1}(p_j)$ is equivalent to the intersection of $Q_1 \cap Q_2 \cap Q_3$
which is thereby proven to be equal to $C$.
Now remains the case, where there is a point $P_0 \in \PSP^3$ for which the lines in
$\theta^{-1}(p)$ form a full pencil, each line $L(p)$ containing $P_0$ and two lines $L(p_1), L(p_2)$ always
meeting in $P_0$.
But this would mean, that $\theta:\PSP^3-C-P_0 \to \PSP_2$ is on $U = \PSP^3-C-P_0$ identical
with the projection from $P_0$ to $\PSP^2$ given by three linear independent linear forms
$$l_1,l_2,l_3 \in H^0(\PSP^3,\Ohol_{\PSP^3}(1)).$$
So we had that the $2 x 2$-minors of
$
\begin{pmatrix}
Q_1 & Q_2 & Q_3 \\\\ l_1 & l_2 & l_3
\end{pmatrix}
$
vanish on the affine cone over $\PSP^3$ and therefore, there is a linear form $l$, such that
$Q_i = l \cdot l_i$. This is obviously impossible, as it would imply, that $C$ lies in the union of
two planes in $\PSP^3$.
So we have excluded the case that $Q_1 \cap Q_2 \cap Q_3$ contains beside $C$ an extra point $P$. Note, that this would also easily follow from  excess intersection theory, see Proposition 13.2 in "3264 and all that" by Eisenbud and Harris.
