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A generic curve over $\mathbb{C}$ of large genus, say at least $24$, will not be a fibre of a Lefschetz pencil or even a hyperplane section.

The reason is that by the theorems of Harris-Mumford and Eisenbud-Harris, the moduli space of curves of large genus is of general type, so there can be no rational curve passing through a generic point. To use this one needs to know that all the smooth fibres of the pencil are not isomorphic. Since the local monodromy around a singular fibre is infinite by the Picard-Lefschetz formula, it suffices to show that there must be at least one singular fibre. But if all fibres are smooth, then $\tilde{Y}$ (the total space of the Lefschetz pencil) must be isomorphic to $C \times \mathbb{P}^1$. This cannot happen since $C \times \mathbb{P}^1$ is not a blow up of any other surface.

Over a field of characteristic zero any very ample linear system contains a Lefschetz pencil, so it follows from the above that the smooth members of the linear system cannot all be isomorphic. But this would give rise to a unirational variety containing a generic point of the moduli space and this is not possible.

A similar argument should work for other classes of varieties whose moduli spaces are not uniruled.

2 added 8 characters in body

A generic curve of large genus, say at least $24$, will not be a fibre of a Lefschetz pencil.

The reason is that by the theorems of Harris-Mumford and Eisenbud-Harris, the moduli space of curves of large genus is of general type, so there can be no rational curve passing through a generic point. To use this one needs to know that all the smooth fibres of the pencil are not isomorphic. Since the local monodromy around a singular fibre is infinite by the Picard-Lefschetz formula, it suffices to show that there must be at least one singular fibre. But if all fibres are smooth, then $\tilde{Y}$ (the total space of the Lefschetz pencil) must be isomorphic to $C \times \mathbb{P}^1$. This cannot happen since $C \times \mathbb{P}^1$ is not a blow up of any other surface.

A similar argument should work for other classes of varieties whose moduli spaces are of general typenot uniruled.

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A generic curve of large genus, say at least $24$, will not be a fibre of a Lefschetz pencil.

The reason is that by the theorems of Harris-Mumford and Eisenbud-Harris, the moduli space of curves of large genus is of general type, so there can be no rational curve passing through a generic point. To use this one needs to know that all the smooth fibres of the pencil are not isomorphic. Since the local monodromy around a singular fibre is infinite by the Picard-Lefschetz formula, it suffices to show that there must be at least one singular fibre. But if all fibres are smooth, then $\tilde{Y}$ (the total space of the Lefschetz pencil) must be isomorphic to $C \times \mathbb{P}^1$. This cannot happen since $C \times \mathbb{P}^1$ is not a blow up of any other surface.

A similar argument should work for other varieties whose moduli spaces are of general type.