Shortest distance from z-axis to given lines The shortest distance between the lines $x+y+2z-3=2x+3y+4z-4=0$ and the z-axis is?. $$\text {Attempt} $$. I first took the cross product of the vectors of two lines ie $i+j+2k,2i+3j+4k $ to get the perpendicular  vector as $-2i+k $. I know that distance between two skew lines is $|\frac {(a_2-a_1). (b_1×b_2)}{(b_1×b_2)}| $. Where $a_2,a_1$ are position vectors from lines and $b_1,b_2$ are the vectors $-2i+k,k (z-axis) $.But how to find the position vectors ?.Thank you.

Let $z=t$.
Hence, $x+y=3-2t$ and $2x+3y=4-4t$, which gives $x=5-2t$ and $y=-2$.
Thus, $(5-2t,-2,t)$ is our line. 
Now, let $(5-2t,-2,t)\subset\pi$ and $(0,0,k)||\pi$.
Let $\vec{n}(a,b,c)$ is a normal of $\pi$.
Thus, $(a,b,c)(0,0,1)=0$ and $(a,b,c)(-2,0,1)=0$ or $c=0$ and $-2a+c=0$, 
which gives that $\vec{n}(0,1,0)$ and the equation of $\pi$ it's
$y=-2$.
Id est, the distance is $2$.
