Construct the intersection of a cube by a plane through $3$ points on its edges, no pair of which is on the same face

Let $PQR$ be the three points defining the plane, belonging to edges $AE$, $BC$ and $FH$ of the given cube (see picture below). Draw from $Q$ a parallel to $BD$, to meet $DF$ at $Q'$. Let $Q''$ be the symmetric of $Q'$ with respect to the midpoint of $ER$ and let $Q'''$ be the symmetric of $Q$ with respect to the midpoint of $PR$. Then $Q'$ and $Q''$ are the projections of $Q$ and $Q'''$ on plane $EDFH$, so that $QQ'''$ and $Q'Q''$ meet at a point $S'$ belonging both to plane $EDFH$ and to plane $PQRQ'''$. Line $RS'$ is then the intersection of those two planes and it intersects edge $EH$ at a point $S$, which is the first of the three points we must construct.

You could now repeat the same construction starting from $P$ and then from $R$, to get other two points $T$ and $U$ on the edges of the cube, but there is a shortcut: just draw from $Q$ a line parallel to $RS$ and another parallel to $PS$, so that $U$ and $T$ will be the intersections of those lines with $AB$ and $CF$ respectively. Finally join $RSPUQT$ to get a hexagon which is the desired intersection.