Non-intersecting lines Given the equations of two lines I have to find the parameter $k$ so the lines don't interesect:
\begin{gather*}
L_1: \left\{(x,y,z)\in \Bbb{R}^3 \,\middle|\, x+2=\frac{3-y}{2}=\frac{z-1}{2} \right\}\\
L_2: \{(x,y,z)\in \Bbb{R}^3 \mid x=a-2k,\ y=2a+k,\ z=4+2k,\ k\in \Bbb{R} \}
\end{gather*}
I am trying to equalize the direction vectors as $(1,-2,2) = k(-2,1,2)$ so the lines are parallels, but I still can't get the value of $k$, any hints?
 A: If you parametrise the two lines so that $\mathbf{a}$ is the direction vector of $L_1$, $\mathbf{b}$ is the direction vector of $L_2$, and $\mathbf{c}$ is a vector from a point on $L_1$ to a point on $L_2$, then the two lines intersect if and only the minimum distance between the lines is zero,
$$\frac{\left\lvert \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \right\rvert}{\left\lVert \mathbf{a} \times \mathbf{b} \right\rVert} = 0$$
(See e.g. Wolfram Mathworld Line-Line Distance article for details and references; the Wikipedia Skew lines' Distance section is less useful.)
As weird as it sounds, it does not matter which points on the two lines you pick for vector $\mathbf{c}$; this is because the minimum distance happens on a vector that is perpendicular to both lines. So, just pick a point on each line you find easy to verify are on their respective lines.
Since you've already shown that $\left\lVert \mathbf{a} \times \mathbf{b} \right\rVert \ne 0$ (because the two lines are not parallel with any value of $k$), the above simplifies to
$$\mathbf{c} \cdot \left( \mathbf{a} \times \mathbf{b} \right) = 0$$
Obviously, this yields the values of $k$ that are not allowed.
A: First of all, let us re-write the equation of $(L_1)$ under the canonical form:
$$x+2=\frac{y-3}{-2}=\frac{z-1}{2}\color{red}{=p} $$
i.e.,
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-2\\3\\1\end{pmatrix}+p\underbrace{\begin{pmatrix}1\\-2\\2 \end{pmatrix}}_U \tag{1}$$
Then, a major remark is that the second set of equations $(L_2)$  doesn't refer to a single line but a whole (affine) plane,
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\4\end{pmatrix}+a\underbrace{\begin{pmatrix}1\\2\\0\end{pmatrix}}_V+k\underbrace{\begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}}_W\tag{2}$$
with normal vector :
$$V \times W = \begin{pmatrix}1\\2\\0\end{pmatrix} \times \begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}=\begin{pmatrix}4\\-2\\5 \end{pmatrix} \ \text{not prop. to} \  U=\begin{pmatrix}1\\-2\\2 \end{pmatrix}$$
($U$ has been defined in (1)).
Therefore, $L_1$ intersects the plane, meaning that there is a value  $k_0$ of $k$ such that (2) is fulfilled.
Set apart this value, in all other cases, there are no solutions.
Remark: the value $k_0$ of $k$ is obtained by solving the linear system obtained by equating (1) and (2):
$$\begin{pmatrix}-2\\3\\1\end{pmatrix}+p\begin{pmatrix}1\\-2\\2 \end{pmatrix}=\begin{pmatrix}0\\0\\4\end{pmatrix}+a\begin{pmatrix}1\\2\\0\end{pmatrix}+k\begin{pmatrix}-2\\ \ \ 1\\ \ \ 2\end{pmatrix}$$
giving the unique solution:

$k=\dfrac{1}{9}$

for the unique value $a=- \ \dfrac{1}{6}.$
