Consider on the first projection plane two points $A$ and $B$. Construct a transversal line $a$ and $b$ through each of these points such that:

1) the angle between the two lines is $30^{\circ}$,

2) the angle between the first projections of the lines is also $30^{\circ}$,

3) the angle between the planes of the lines is $60^{\circ}$.
The first traces of the lines are points $A$ and $B$, so the first trace line of the plane is the line $|A B|$. The first dihedral angle of the plane is $90^{\circ}$, which means the sought lines can lie in two planes: $S_{1}$ and $S_{2}$. Since $\widehat{a b}=\widehat{a^{\prime} b^{\prime}}=30^{\circ}$, the intersection points of the rotated lines and their images $(C)_{1}$ and $C^{\prime}$ lie on a circle whose chord is $A B$ and central angle is $60^{\circ}$. Due to the $60^{\circ}$ first dihedral angle, $\overline{s_{1} C^{\prime \prime}}: \overline{s_{1}(C)_{1}}=1: 2$. Since the ray parallel to $s_{1}$ bisects the chord $\overline{C^{\prime}(C)_{1}}$, $\overline{C^{\prime} s_{1}}=\frac{2}{3} \overline{O s_{1}}$, where $O$ is the center of the circle. The number of solutions is 8.

George Bizám (Bólyai g. VI. r. o. Budapest).