Will two planes that are perpendicular to the same plane necessarily be parallel?
Consider the cube $A B C D A 1 B 1 C 1 D 1$. Each of the planes of the faces $A B B 1 A 1$ and $B B 1 C 1 C$ is perpendicular to the plane $A B C D$, since the corresponding dihedral angles are $90^{\circ}$. At the same time, the planes $A B B 1 A 1$ and $B B 1 C 1 C$ intersect along the line $B B 1$.

## Answer

No.

Let $A$ be a point in space, $A 1$ be the projection of point $A$ onto the plane $\alpha, A A 1=a$. Through point $A$ passes a plane that forms an angle $\phi$ with the plane $\alpha$ and intersects the plane $\alpha$ along the line $l$. Find the distance from point $A 1$ to the line $l$.

## Solution

If $\phi=90^{\circ}$, then the line $l$ passes through point $A 1$. In this case, the desired distance is 0. Let the plane $\beta$, passing through the given point $A$, form an angle $\phi \neq 90^{\circ}$ with the plane $\alpha$. Drop a perpendicular $A 1 B$ from point $A 1$ to the line $l$. Then the distance from point $A 1$ to the line $l$ is equal to the length of the segment $A 1 B$. By the theorem of three perpendiculars, $A B \perp l$. Therefore, $A B A 1$ is the linear angle of the dihedral angle formed by the planes $\alpha$ and $\beta$. By the problem's condition, $\angle A B A 1=\phi$. From the right triangle $A B A 1$, we find that

$$
A 1 B=A A 1 \cdot \operatorname{ctg} \angle A B A 1=a \operatorname{ctg} \phi
$$

## Answer

$\operatorname{actg} \phi$.