How to show that three lines are all crossed by infinitely many lines at the same time

Question

the question is here for clarity

If three lines l_1, l_2, l_3 are given as l_1={(s,0,1)|s an element of R} and l_2={(1,t,0)|t an element of R} and l_3={(0,1,u)|u an element of R} then show that there are infinitely many lines that cross l_1,l_2 and l_3 at the same time.

I tried to separate the given equations of the lines like l_1=(0,0,1)+s(1,0,0), l_2=(1,0,0)+t(0,1,0) and l_3=(0,1,0)+u(0,0,1) but it turns out that from the above equations, the direction vectors of these three lines are perpendicular to one another and the direction vectors turned out to be the standard i, j and k. I thought the three lines would be all on the same plane and not perpendicular to one another. Is there something that I am missing? Any hint or a reference to a book with similar problems will really be appreciated. Please...

Answer 1

Let $\ell_1$, $\ell_2$ be not sets of points but arbitrary points of that sets.
We construct the desired line as passing through $\ell_1$, $\ell_2$, i.e. $\{(s,0,1)+v((1,t,0)-(s,0,1))\}$.
Then make it intersect with $\ell_3$:
$$\begin{cases}s+v(1-s)=0,\\ 0+v(t-0)=1,\\ 1+v(0-1)=u\end{cases}$$ $$\begin{cases} s\ne 0,\\ t = \frac{s - 1}{s}, \\s - 1\ne 0, \\u = \frac{1}{1 - s}, \\v = \frac{s}{s - 1} \end{cases}$$ and show that there are infinitely many solutions of the intersection set of equations, which differs not only by the scale of direction vector (as we will have $3$ equations with $4$ variables we need to show that the system is solvable and have at least $1$ free variable).
The lines go through points $(s,0,1)\in\ell_1, (1,\frac{s - 1}{s},0)\in\ell_2, (0,1,\frac{1}{1 - s})\in\ell_3$ and none of them are the same line, so there's infinity of lines, QED.
The same as this answer with $x+1=s$.

Answer 2

Three distinct points $A, B, C$ in $\mathbb R^3$ are on the same line $\iff \exists x \in \mathbb R$ such that $A-B = x(B-C)$. Setting $A=(s,0,1), B=(1,t,0), C=(0,1,u)$ this translates to: $$s - 1 = x$$ $$-t = x(t-1)$$ $$1 = -xu$$

Solving for $s, t$ and $u$ in terms of $x$, we get that the points $(x+1,0,1), (1,\frac{x}{x+1},0)$ and $(0,1,-\frac{1}{x})$ are on the same line and also on $l_1, l_2$ and $l_3$, respectively. A moment's thought will convince you that the lines determined by all the possible choices of $x$ are all different.