I would like to make one guess concerning one particular situation.

Correction In fact, the following guess is wrong (thanks to jvp)! But I don't want to delete this answer, since jvp left a comment to it with a cool link, that shows why the guess is wrong

Guess. Suppose that $M\cong \mathbb CP^1$, and $E=O(1)\oplus...\oplus O(1)$. In this case $X$ is rational, i.e. birational to $\mathbb CP^n$.

I have to say that I don't know how to prove this guess and even not 100% sure that this guess is correct. But first of all, this statement is obviously true if $n=2$ and I rather belive it should be possible to prove this for $n=3$ using twistor theory. The reason to make this guess is that if you have such a line in $X$ you can start to deform it and it deforms (at least locally) in exactly the same way as a line in $\mathbb CP^n$. Also, it is obvious that $X$ is rationally connected in this case.

Of course, this guess is a pure speculation. But maybe someone can prove it? :)

Remark 1. You can always blow up something in $X$ that does not touch the image of the zero section of $N$, so one can only say something (if one can at all) about birational type of $X$.

Remark 2. Note that if you don't assume $X$ to be Kaehler, but only ask it to be complex, then the guess is completely wrong. There exist huge amount of complex 3-folds containing lines with neighborhoods biholomorphic to a neighbourhood of a line in $\mathbb CP^3$. Such examples come from twistor theory -- one should take the twistor space of a conformally flat 4-manifold.

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I would like to make one guess concerning one particular situation.

Guess. Suppose that $M\cong \mathbb CP^1$, and $E=O(1)\oplus...\oplus O(1)$. In this case $X$ is rational, i.e. birational to $\mathbb CP^n$.

I have to say that I don't know how to prove this guess and even not 100% sure that this guess is correct. But first of all, this statement is obviously true if $n=2$ and I rather belive it should be possible to prove this for $n=3$ using twistor theory. The reason to make this guess is that if you have such a line in $X$ you can start to deform it and it deforms (at least locally) in exactly the same way as a line in $\mathbb CP^n$. Also, it is obvious that $X$ is rationally connected in this case.

Of course, this guess is a pure speculation. But maybe someone can prove it? :)

Remark 1. You can always blow up something in $X$ that does not touch the image of the zero section of $N$, so one can only say something (if one can at all) about birational type of $X$.

Remark 2. Note that if you don't assume $X$ to be Kaehler, but only ask it to be complex, then the guess is completely wrong. There exist huge amount of complex 3-folds containing lines with neighborhoods biholomorphic to a neighbourhood of a line in $\mathbb CP^3$. Such examples come from twistor theory -- one should take the twistor space of a conformally flat 4-manifold.