The vector equation of two lines are

$$\begin{aligned} & \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\ & \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k} \end{aligned}$$

Then the shortest distance between them is

The place $$x-2 y+z=0$$ is parallel to the line

The lines $$\frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$$ and $$\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$$ are perpendicular to each other, then $$a$$ equals to

If two lines $$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$$ and $$L_2: \frac{x-3}{1}=\frac{y-k}{2}=z$$ intersect at a point, then $$2 k$$ is equal to