3-4. Given positive numbers \(h, s_{1}, s_{2}\) and a triangle \(A B C\) located in space. In how many ways can a point \(D\) be chosen so that in the tetrahedron \(A B C D\), the height dropped from vertex \(D\) is equal to \(h\), and the areas of the faces \(A C D\) and \(B C D\) are \(s_{1}\) and \(s_{2}\) respectively (consider all possible cases)
Solution 4. Answer: \(0, 2, 4\) or 8.

To ensure that the height dropped from vertex \(D\) is equal to \(h\), point \(D\) must lie in one of two planes \(\Pi_{1}\) and \(\Pi_{2}\), parallel to the plane \(A B C\). To ensure that the area of face \(A C D\) is equal to \(s_{1}\), point \(D\) must lie on a cylinder with axis \(A C\), and to ensure that the area of face \(B C D\) is equal to \(s_{2}\), point \(D\) must lie on a cylinder with axis \(B C\). The intersection of plane \(\Pi_{1}\) with the first cylinder is either a pair of lines, a single line, or an empty set, and the lines must be parallel to \(A C\). For the second cylinder, the lines obtained are parallel to line \(B C\), which intersects line \(A C\). Therefore, when the cylinders are intersected by plane \(\Pi_{1}\), we get either an empty set, a pair of intersecting lines, a line intersecting a pair of parallel lines, or a pair of parallel lines intersecting another pair of parallel lines. The number of points belonging to both cylinders and the plane is \(0, 1, 2\), and 4, respectively. We get the same number of intersection points for plane \(\Pi_{2}\).