Finding a line through 4 other lines! This one's probably easy, but I'm dreadfully stuck and can't seem to figure out a decent method.
I have the following lines:
$$a: \vec{x}(\lambda)= \left( \begin{array}{ccc}
4  \\
-2  \\
-2 \end{array} \right) + \lambda\left( \begin{array}{ccc}
1  \\
-1  \\
-1 \end{array} \right) 
$$
$$b: \vec{x}(\mu)= \left( \begin{array}{ccc}
-1  \\
1  \\
-3 \end{array} \right) + \mu\left( \begin{array}{ccc}
1  \\
0  \\
2 \end{array} \right) 
$$
$$c: \vec{x}(\nu)= \left( \begin{array}{ccc}
1  \\
0  \\
5 \end{array} \right) + \nu\left( \begin{array}{ccc}
0  \\
-2  \\
1 \end{array} \right) 
$$
$$d: \vec{x}(\tau)= \left( \begin{array}{ccc}
3 \\
-2  \\
0 \end{array} \right) + \tau\left( \begin{array}{ccc}
-1  \\
1  \\
1 \end{array} \right) 
$$
I have to find the line that intersects all four of these lines. How do I go about doing this? 
Cheers!
 A: Note that $a$ and $d$ are parallel, so whatever line it is, it needs to lie in the plane containing $a$ and $d$. Considering it also needs to intersect $b$ and $c$, try figuring out which two points those two lines intersect the $ad$-plane.
Edit: Here's a full answer.
$ad$-plane: The $ad$-plane's normal vector is orthogonal to $(-1, 1, 1)$ as well as $\vec{a}(0) - \vec{d}(0) = (1, 0, -2)$. We therefore have a normal vector given by
$$
(-1, 1, 1)\times (1, 0, -2) \\
= (1\cdot(-2) - 1\cdot 0, 1\cdot 1 - (-1)\cdot (-2), (-1)\cdot 0 - 
1\cdot 1)\\
= (-2, -1, -1)
$$
I elect to choose the negative of this vector, for estethic reasons.
Inserting $\vec{d}(0)$ into the general equation for a plane, we have:
$$
2\cdot 3 + 1\cdot (-2) + 1 \cdot 0 = 4
$$
and therefore the $ad$-plane is given by $2x + y + z = 4$.
$b$-intersection: The $\mu$ for the point where the $b$-line intersects the $ad$-plane is given by
$$
2(-1 + \mu) + 1 -3 + 2\mu = 4\\
4\mu = 8\\
\mu = 2
$$
so the intersection point is $B = \vec{b}(2) = (1, 1, 1)$.
$c$-intersection: The $\nu$ for the point where the $c$-line intersects the $ad$-plane is given by
$$
2\cdot 1 -2\nu + 5 + \nu = 4\\
-\nu = -3\\
\nu = 3
$$
so the intersection point is $C = \vec{c}(3) = (1, -6, 8)$.
The line: We need the line $\vec{l}(\gamma)$ that goes from $B$ to $C$. It is given by
$$
\vec{l}'(\gamma) = B + \gamma(C - B)\\
= (1, 1, 1) + \gamma(0, -7, 7)
$$
which I would like to rewrite to:
$$
\vec{l}(\gamma) = (1, 1, 1) + \gamma(0, -1, 1)
$$
For reference, the four intersection points are:


*

*$al$: $\lambda = -3, \gamma = 0, (1, 1, 1)$

*$bl$: $\mu = 2, \gamma = 0, (1, 1, 1)$

*$cl$: $\nu = 3, \gamma = 7, (1, -6, 8)$

*$dl$: $\tau = 2, \gamma = 1, (1, 0, 2)$

