1.10. At the base of a regular triangular prism lies a triangle $ABC$ with side $a$. Points $A_{1}, B_{1}$, and $C_{1}$ are taken on the lateral edges, the distances from which to the plane of the base are $a / 2$, $a$, and $3 a / 2$. Find the angle between the planes $ABC$ and $A_{1} B_{1} C_{1}$.

## § 3. Lines forming equal angles with lines and planes
1.10. Let $O$ be the point of intersection of the lines $A B$ and $A_{1} B_{1}$, $M$ be the point of intersection of the lines $A C$ and $A_{1} C_{1}$. First, we will prove that $M O \perp O A$. For this, we will take points $B_{2}$ and $C_{2}$ on the segments $B B_{1}$ and $C C_{1}$ such that $B B_{3}=C C_{2}=A A_{1}$. It is clear that $M A: A A_{1}=$ $=A C: C_{1} C_{2}=1$ and $O A: A A_{1}=A B: B_{1} B_{2}=2$. Therefore, $M A: O A=1: 2$. Moreover, $\angle M A O=60^{\circ}$, which means $\angle O M A=90^{\circ}$. Therefore, the plane $A M A_{i}$ is perpendicular to the line $M O$, along which the planes $A B C$ and $A_{1} B_{1} C_{1}$ intersect. Hence, the angle between these planes is equal to the angle $A M A_{1}$, which is $45^{\circ}$.