This is a statement in Harris A first course of AG's Thm 3.12.
Let $X$ be a subvariety of $Y\times P^2$ where $Y$ is a variety of $A^n$ and $P^2$ is 2 dimensional projective space. Choose any $p\in P^2$ such that $X\cap(Y\times\{p\})\neq\phi$. Then $V=\{q\in Y\vert (q,p)\not\in X\}$ is closed.
I was trying to see $V$ being closed. I tried $Y-V=\{q\in Y\vert (q,p)\in X\}=X\cap(Y\times \{q\})$ which should be closed in $Y$ as I can fix the generator's $P^2$ variable at p which should yield a polynomial vanishing exactly on $Y-V$. I think I am wrong here as I did not get $V$ being closed.
Don't have it with me, but I think this must be a typo. Consider $X \subset \mathbb{A}^3\times \mathbb{P}^2$ given by $ax + by + cz = 0$ ( where $x,y,z$ are homogeneous coordinates for $\mathbb{P}^2$ ). Points on this variety $(p,q)$ are points where $p$ is a point on the plane perpendicular to the line through the origin corresponding to $q$. The set you described $V$ then is $\mathbb{A}^3$ minus a plane, which is clearly not closed.
If you replace the $\notin$ with $\in$, then $V$ is closed by the main theorem of elimination theory ( it's a projection to $Y$ of a closed fiber over a point in $\mathbb{P}^2$ )