Bariz - Problem

Draw a line $\ell''$ through $A$ parallel to $\ell$ and let $XZ$ be the common perpendicular of $\ell$ and $\ell''$ with $Z\in\ell''$. Then $Y$ lies on the circle with the diameter $XZ$ which lies in the plane passing through $X$ and perpendicular to $\ell$. This means that $Y$ lies on the cylinder with opposite generetrices $\ell$ and $\ell''$. It's not hard to see that any point of this cylinder serves an endpoint of common perpendicular for $\ell$ and some line through $A$. Hence the locus is the aforementioned cylinder.