1. In space, there are $n(n \geqslant 3)$ planes, among which any three planes do not have a common perpendicular plane. Then, among the following four conclusions:
（1）No two planes are parallel to each other;
（2）No three planes intersect in a single line;
(3) Any two lines of intersection between planes are not parallel:
(4) Each line of intersection between planes is intersected by $n-2$ planes.

Among them, the number of correct conclusions is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
- 1. (D).

If two planes $\alpha_{1}, \alpha_{2}$ are parallel, then the common perpendicular plane of $\alpha_{1}, \alpha_{3}$ is also a perpendicular plane of $\alpha_{2}$, which contradicts the condition that there is no common perpendicular plane with any plane. Conclusion (1) is true.

If three planes intersect at one line, then the perpendicular plane of this line is the common perpendicular plane of the three planes, which contradicts the given condition. Conclusion (2) is true.

If two intersection lines are parallel, then the perpendicular plane of these two lines is at least the common perpendicular plane of three planes, which contradicts the given condition. Conclusion (3) is true.

If the intersection line $l$ of $\alpha_{1}$ and $\alpha_{2}$ does not intersect with plane $\alpha_{3}$, there are two possibilities: $l$ lies on $\alpha_{3}$ (i.e., $\alpha_{1}, \alpha_{2}, \alpha_{3}$ intersect at $l$), or $l$ is parallel to $\alpha_{3}$ (i.e., $\alpha_{3}$ is parallel to either $\alpha_{1}$ or $\alpha_{2}$, or $\alpha_{3}$ intersects $\alpha_{1}, \alpha_{2}$ at three parallel lines). Both of these possibilities contradict the given conditions. Conclusion (4) is true.