How to prove that there can be infinite lines intersecting 3 skewed lines? Few days back,I learnt somewhere in Mathematics Stack Exchange (don't exactly remember the question) that there can be infinite lines intersecting 3 skewed lines.But I'm not able to visualize or prove this particular fact.Any help appreciated...
 A: Let's call the three skew lines $L1, L2, L3$. Take a point $P$ that's on $L1$ but not on $L2$. Draw a line from $P$ to each point of $L2$. This gives you a family of lines $F$ that forms a plane. This plane intersects the line $L3$ somewhere. In other words, there is a line $L$ in the family $F$ that intersects $L3$. This line $L$ intersects all three of the given lines.
If you start with a different $P$, then you'll get a different line $L$, but this $L$ will still have the desired properties.
In fact, in general, you can construct a line that intersects four given lines. For some sketchy details, see answers to this question, and also see Sommerville, Analytical Geometry of Three Dimensions, Section 9.4.
A: I will add part of the setup here. See picture at  https://en.wikipedia.org/wiki/Hyperboloid#Properties  of a hyperboloid of one sheet that is double ruled, meaning there are two families of lines contained in it. This is also the case with the very different hyperbolic paraboloid, the example they give is $z = xy.$ See also https://en.wikipedia.org/wiki/Ruled_surface
Given your three pairwise skew lines, we can make a quadric surface that contains all three, and infinitely many other lines. I did an easy example:
I decided to look into the doubly ruled surface that contains the three parametrized lines,
$$ x=1, \; \;  y = t, \; \;  z = t,  $$
$$ x=-1, \; \;  y = -t, \; \;  z = t,  $$
$$ x=-t, \; \;  y = 1, \; \;  z = mt,  $$
for real constant $m.$ The resulting quadric is given by
$$ \color{blue}{  m x^2 + y^2 - z^2 + (1-m)zx +(m-1)y = m.}  $$
When $m=1,$ we get the surface depicted on Wikipedia several times, $x^2 + y^2 = z^2 + 1.$
I am thinking in terms of $m > 0,$ although it appears $m=0$ gives one of those paraboloids. 
I will try to find all straight lines contained in the surface $$ \color{blue}{  m x^2 + y^2 - z^2 + (1-m)zx +(m-1)y = m.}  $$ This is the kind of thing the algebraic geometry people can do in their sleep, but I've never tried it before. 
EDIT: wasn't that bad. Given hyperboloid of one sheet $u^2 + v^2 - w^2 = (m+1)^2,$ through the point $((m+1) \cos \theta, (m+1) \sin \theta,0)$ the two parametrized lines for $(u,v,w)$ are
$$  ((m+1) \cos \theta - t \sin \theta, (m+1) \sin \theta + t \cos \theta, t), $$
$$  ( (m+1) \cos \theta + t \sin \theta, (m+1) \sin \theta - t \cos \theta, t). $$
In either case solve for $x,y,z$ using
$$ u = (m+1)x,  \; \; v = 2y+m-1, \; \; w = (m-1)x + 2z $$
