The picture shows a 2-dimensional representation of a 3-dimensional tetrahedron.
The triangle ABC forms the base and O is a common vertex of the tetrahedron above the plane of the triangle ABC. If we drop a perpendicular from vertex O onto the plane of the triangle ABC it meets this plane at point O'.
The red lines are meant to represent lines coming out of the 2D plane of ABC into the third dimension. The sides of the tetrahedron form three triangles with a common height h (the length of the perpendicular): OAB, OBC, OAC. There are also 3 right triangles formed from the perpendicular OO' and the vertices of the base: OO'A, OO'B, OO'C. These have a common leg OO' and hypotenuses of length a, b and c. Each right triangle can be rotated by keeping OO' as a fixed hinge and swinging each vertex, A, B or C, without altering the length of the hypotenuse, but in doing this the shape of the triangular base ABC changes. Therefore the length of AB changes. The positioning of P and Q as described in the question doesn’t appear to affect AB, so there is some information missing that will fix AB so that it can’t be changed. What information is missing? For example, is P in the plane of the base; is it perpendicular to it; is it on OA extended; is it on AB extended?
Take vertex A. The point P can be anywhere on the surface of a sphere centre at A and radius=a. The sphere can intersect the tetrahedron at any one of various places, any of which could be point P. Similarly for vertex B: the point Q can be anywhere on the surface of a sphere centre at B and radius=2b. The line PQ joins points on the surfaces of the two spheres. That gives us many possible lines. ???