In the plane $s$, find a point which is at a distance $m$ from a given line $a$ and from a given plane $s_{1}$.
Solution. The sought point lies on three geometric places: one geometric place is the given plane $s$; the other is a cylinder whose axis is the line $a$ and the radius of its base circle is $m$; the third is a plane $s_{2}$, which is at a distance $m$ from the plane $s_{1}$ and is parallel to it.

Accordingly, the point of intersection of the line of intersection of the planes $s$ and $s_{2}$ with the cylinder must be found. Since there are two possible $s_{2}$ planes, the number of solutions is generally 4.

To simplify the construction, the line $a$ should be transformed twice to become a projecting line, at which point the intersection of the cylinder with the fourth image plane is a circle, and thus the fourth projections of the sought points can be directly obtained as the common points of the line of intersection of the planes $s$ and $s_{2}$ and the base circle of the cylinder. From here, the other projections of the points can also be constructed.

(Szántó László, Pécs.)

The problem was also solved by: Ehrenfeld N. and Lengyel M.