The following question makes sense in a more general setting but for sake of simplicity let me stick to a particular case.


Consider the degree three Veronese embedding $V\subset\mathbb{P}^9$ of $\mathbb{P}^2$ via the complete linear system of plane cubics.


Does there exist a point $p\in\mathbb{P}^9$ not lying on any secant line to $V$ (properly secant, if $p$ is on a tangent line it is ok), but lying on infinitely many planes that intersects $V$ in three points in linear general position?



Linear spaces secant to Veronese varieties
Here is an answer in terms of power sum decompositions of polynomials. A point $p \in \mathbb{P}^9$ corresponds to a homogeneous polynomial $P$ of degree $3$ in $3$ variable, defining a plane cubic. Points of the Veronese $q \in V$ correspond to pure powers of linear forms, $q = \ell^3$. A point $p$ lies in the span of points $q\_1,\dotsc,q\_r \in V$ if and only if the corresponding polynomial $P$ is a linear combination of powers, $P = c\_1 \ell\_1^3 + \dotsc + c\_r \ell\_r^3$, where the $c\_i$ are scalars. Such an expression is often called a power sum decomposition. The least number of terms in any power sum decomposition of a polynomial is called the polynomial's Waring rank. In these terms you are asking whether there exists a plane cubic of rank $3$, with infinitely many power sum decompositions in three terms.


The polynomial $x^2 y$ has this property (assuming the characteristic is not $2$ or $3$). A power sum decomposition with $3$ terms is given by
$$
 x^2 y = \frac{1}{6}(y+x)^3 + \frac{1}{6}(y-x)^3 - \frac{2}{6} y^3 .
$$
More generally,
$$
 x^2 y = \frac{1}{6a^2}(y+ax)^3 + \frac{1}{6a^2}(y-ax)^3 - \frac{2}{6a^2} y^3 ,
$$
giving infinitely many power decompositions. In your terms, the planes $\operatorname{span}\{(y+ax)^3, (y-ax)^3, y^3\}$ are distinct, and meet $V$ at those three points, which are linearly independent. Those aren't all of the decompositions. I can try to write the rest if you need. But there are infinitely many, anyway.


To see that the planes in this family are pairwise distinct, note that $(y+bx)^3$ can't lie in the span of $(y+ax)^3,(y-ax)^3,y^3$ unless $b \in \{a,-a,0\}$, because the matrix whose rows are the coefficients of those cubic forms is a Vandermonde matrix, so they are linearly independent unless $b$ has one of those values.


The polynomial $x^2 y$ does lie on a tangent line to the Veronese:
$$
 x^2 y = \lim\_{t \to 0} \frac{ (x+ty)^3 - x^3 }{ 3t } ,
$$
a limit of secant lines through $(x+ty)^3$ and $x^3$ (a tangent line through $x^3$).


But $x^2 y$ doesn't lie on any proper secant line to $V$, meaning we can't write $x^2 y$ as $\ell\_1^3 + \ell\_2^3$. The first one factors with a repeated factor, while the second one factors as
$$
 (\ell\_1 + \ell\_2)(\ell\_1 + \omega \ell\_2)(\ell\_1 + \omega^2 \ell\_2),
$$
$\omega$ a $3$rd root of unity, which has distinct factors.


**Edit:** The other decompositions of $x^2 y$ actually aren't that hard to describe. Let $B$ (for "binary") be the span of $x^3,x^2 y, x y^2, y^3$ in $\mathbb{P}^9$, so $B$ is a $\mathbb{P}^3$, parametrizing the binary cubics in the variables $x,y$. The intersection $B \cap V$ is a twisted cubic, the pure powers $(ax+by)^3$. Now the point is that any $2$-plane passing through $x^2 y$ and contained in $B$ will cut that twisted cubic in three points, giving a three-term power sum decomposition.


And the converse holds: any three-term power sum decomposition of $x^2 y$ doesn't involve $z$, that is, the three terms are all linear forms in just $x$ and $y$. It's not too hard to prove this "hands-on" for this case. More generally, Buczyński and Landsberg write about this in terms of "rank preserving pairs", <https://arxiv.org/abs/0909.4262>.


In a comment it is observed that $x^2 y$ depends essentially on only $2$ variables in the sense that its second derivatives span a $2$ dimensional subspace of the linear forms, and there is the question whether there is a form $P$ with the same rank properties (rank $3$ with infinitely many rank decompositions) but depending essentially on all variables ("concise"). The answer is no. Ranks of plane cubics have been worked out by many authors (it's been a bit of a running example, used to demonstrate new approaches) including B. Segre, Reznick, Comon and Mourrain, Kleppe, Landsberg and myself, plenty of others. The only plane cubics of rank $3$ are (up to linear change of coordinates) $x^2 y$ and $x^3 + y^3 + z^3$. The derivatives of $P = x^3 + y^3 + z^3$ do span all the linear forms, and it has rank $3$. But it doesn't have infinitely many rank decompositions. In fact the rank decomposition is unique.


Briefly, suppose $x^3 + y^3 + z^3 = \ell\_1^3 + \ell\_2^3 + \ell\_3^3$. Taking first derivatives, each of $x^2$, $y^2$, $z^2$ lies in the span of $\{\ell\_1^2,\ell\_2^2,\ell\_3^2\}$. In fact $\{x^2,y^2,z^2\}$ must have the same span. So each $\ell\_i^2 = a x^2 + b y^2 + c z^2$. The only such squares are just scalar multiples of the $x^2,y^2,z^2$ (i.e., two of the coefficients $a,b,c$ vanish). So the $\ell\_i$ are the same as $x,y,z$, up to order and scalar multiple.


For more on homogeneous polynomials with infinitely many rank decompositions, look for keywords like "identifiable polynomials" (or non-identifiable, really). Good luck!


