3. In space, there are $n(n \geqslant 3)$ planes, among which any three planes do not have a common perpendicular plane. There are 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 intersects with $n-2$ planes.

Among these, the number of correct conclusions is . $\qquad$
3. 4 .

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

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

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

If the line of intersection $l$ of $\alpha_{1}$ and $\alpha_{2}$ does not intersect with plane $\alpha_{3}$, then there are two possibilities: $l$ lies on $\alpha_{3}$ (where $\alpha_{1}, \alpha_{2}, \alpha_{3}$ intersect at $l$), or $l$ is parallel to $\alpha_{3}$ (where $\alpha_{3}$ 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.