Identify the hyperboloid of one sheet that contains three skew lines on its surface

Question

I claim that three skew lines define a unique hyperboloid of one sheet that contains all of the three lines on its surface.

Suppose you are given three lines in parametric form in $3D$, described as follows

$r_i(t) = P_i + t\ d_i ,\ t \in \mathbb{R},\ i = 1, 2, 3 $

where $P_i$ is a point on the $i$-th line and $d_i$ is the direction vector of the $i$-th line.

Find the equation of the hyperboloid of one sheet that contains all three lines on its surface.

My attempt:

My attempt at this problem is contained in my solution that follows.

My question:

Is it true that three skew-lines define a unique hyperboloid of one sheet that contains them on its surface ? Any hints, remarks, and alternative solutions are appreciated.

Answer 1

I quote "Geometry and the imagination" pp 14-15, Hilbert, David, 1862-1943, author; Cohn-Vossen, S. (Stephan), 1902-1936, author; Nemenyi, P., translator

... the general hyperboloid of one sheet contains two families of straight lines, since a dilation always transforms straight lines into straight lines. Again the lines are arranged in such a way that every line of one family has a point in common with every line of the other family and any two lines of the same family are skew.

This gives rise to the following construction of the hyperboloid of one sheet (see Fig. 21). We start with any three straight lines of one family. Since no two of them have a plane in common, every point $P$ on one of them is on one and only one straight line $p$ meeting the other two given lines, namely the intersection of the plane containing $P$ and the second line with the plane containing $P$ and the third line. $p$ has three points in common with the hyperboloid. But no straight line can intersect a quadric in more than two points. Consequently $p$ must be one of the lines on the hyperboloid. If the point $P$ traverses the first line, the corresponding line $p$ will take on the positions of all the straight lines of that family of the hyperboloid to which the first line does not belong. If we choose any three straight lines of this family, we can get the other family by the same procedure, and of course this will include the three lines with which we started originally. The construction shows that every pair of straight lines of the same family must be skew, provided it is at all possible to find three non-coplanar straight lines in one of the families. For, if we could find three straight lines with which to carry out the construction, while $p$ and $p'$ were to meet at a point $Q$, (Fig. 21), then the original lines would all have to lie in the plane $PP'Q$, in contradiction to one of the assumptions. On the other hand, it is clear that our surface would be not a hyperboloid but a plane, if three lines of the same family always turned out to be coplanar.

Thus three skew straight lines always define a hyperboloid of one sheet, except in the case where they are all parallel to one plane (but not to each other). In this case they determine a new type of second-order surface, called the hyperbolic paraboloid, which does not include any surface of revolution as a special case. ...