Vrftjkry

Let $Xsubset mathbb P^n$ be a smooth hypersurface over base field $mathbb C$. A pencil of hyperplanes is just a projective line $(X_t)$ in $mathbb P^{n*}$. It is called a Lefschetz pencil if it satisfies the following: (I followed the definition in SGA)

(i) The axis (intersection of $X_t$'s) intersects with $X$ transversally.

(ii) Almost all (enough for one) $X_t$ intersects with $X$ transversally.

(iii) The non-transversal intersection in (ii) contains one node.

And on SGA it also proves that almost all pencils are Lefschetz. Now I want to know how many are those non-Lefschetz pencils, more precisely:

Does there exist some $X$, on which all the non-Lefschetz pencils cover the $mathbb P^{n*}$? i.e. for any hyperplane section $H$, we can find non-Lefschetz pencil $(X_t)$ pass through it.

I guess it does not exist, but have no idea how to prove it. Any hints or references would be helpful.

Remark

I should remove the condition (i) otherwise it is trivial.

Assume there is a single hyperplane section $X_0$ of $X$ which has worse singularity than just a node. Then any pencil containing $X_0$ is non-Lefschetz, and these pencils cover all the $mathbb{P}^{n*}$.

Of course, if there is no such $X_0$ as above, this construction does not work. However, I think the only case when such $X_0$ does not exist is that of $X$ being a smooth quadric. In that case the projective dual variety to $X$ is yet another smooth quadric, and lines in $mathbb{P}^{n*}$ tangent to it provide non-Lefschetz pencils that cover all the $mathbb{P}^{n*}$.