Let $X\subset\mathbb{P}^n$ be a smooth projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ if we assume furthermore that $X\cap H$ does not contain ruled components?

EDIT: As @Lev Borisov pointed out in his answer, such examples exist if we allow arbitrary geometry of the intersection set.

Let $X\subset\mathbb{P}^n$ be a smooth projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ if we assume furthermore that $X\cap H$ does not contain ruled components?

EDIT1: As @Lev Borisov pointed out in his answer, such examples exist if we don't put any restrictions on the geometry of the intersection set.

EDIT2: The example can be generalized in such a way that the set $X\cap H$ is almost arbitrary. So the assumption about ruled components does not make any difference.

Let $X\subset\mathbb{P}^n$ be a smooth projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ if we assume furthermore that $X\cap H$ does not contain ruled components?

If we don't require $X$ to be smooth, then there are some trivial examples, say $x^2-yz\subset\mathbb{P}^3$, showing that such a hyperplane can exist. However, it seems that the smoothness constraint makes such examples a little harder (or maybe impossible?) to find.

Let $X\subset\mathbb{P}^n$ be a smooth projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ if we assume furthermore that $X\cap H$ does not contain ruled components?

EDIT: As @Lev Borisov pointed out in his answer, such examples exist if we allow arbitrary geometry of the intersection set.