5 points determine a conic uniquely

This is exercise 3.5.5 in Gathman's notes. We proved so far that zero set of degree $$2$$ homogeneous polynomials in variables $$x_0,x_1,x_2$$ can be identified as $$\mathbb{P}^5$$, and conics can be identified as an open subset $$U\subset\mathbb{P}^5$$. Also, given a point $$P\in\mathbb{P}^2$$, and let $$F: \mathbb{P}^2\mapsto\mathbb{P}^5$$ be the Veronese embedding. Then a conic $$C\in U$$ passing through $$P$$ if and only if $$F(P)\cdot C = 0$$. The last part requires to prove that given $$P_1,\ldots, P_5\in\mathbb{P}^2$$ such that no three of them is on the same line, then there is a unique conic passing through the $$5$$ points.

What I prove so far is that the matrix $$\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$$ is a $$5\times 6$$ matrix, and hence the null set have dimension at least $$1$$ which corresponds to vanishing of degree $$2$$ homogeneous polynomials passing though $$P_1,\ldots, P_5$$. I also showed that if no three of $$P_1,\ldots, P_5$$ lies on the same line, then that degree $$2$$ homogeneous polynomial must be a conic (irreducible).

However, to prove the uniqueness, I need to show that $$F(P_i)$$ are linearly independent, and I got stuck on this. I read answers from another post, and it suggests that linear relation on $$F(P_1),\ldots, F(P_5)$$ can be pulled back to linear relation on $$P_1,\ldots, P_5$$. I don't see why this is true.

You will need the condition that $$P_i$$'s are in general position, which means an open subset (in the Zariski sense) of all the possible configurations $$\{(P_1,\ldots,P_5)\}$$.
Precisely, you need to show the condition that $$\{F(P_i)\}$$ linearly independent is an open condition. Indeed, by basic linear algebra, the dependency of $$\{F(P_i)\}$$ is equivalent to that the determinants $$D_i=D_i(P_1, \ldots, P_5)$$ $$(i=1,\ldots,6)$$ of all the $$5\times 5$$ submatrices of $$\begin{bmatrix}F(P_1)\\\vdots\\F(P_5)\end{bmatrix}$$ are zero, which is a closed condition.