The shortest distance between the z-axis  and  the line of 
intersection of ,   $x+y+2z-3=0$ and $2x+3y+4z-4=0$  is:
 

The equation of any plane containing the given line is  $\left(x+y+2z-3\right)+\lambda \left(2x+3y+4z-4\right)=0$$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\Rightarrow |\overrightarrow{a}+\overrightarrow{b}{|}^2=|\overrightarrow{c}{|}^2=1$
If the plane is parallel to z-axis whose direction cosines are 0, 0, 1; then the normal to the plane will be perpendicular to z-axis 
$\therefore \left(1+2\lambda \right)\left(0\right)+\left(1+3\lambda \right)\left(0\right)+\left(2+4\lambda \right)\left(1\right)=0$ $\Rightarrow \lambda =-\frac{1}{2}$
Put in eq. (1), the required plane is 
$\left(x+y+2z-3\right)-\frac{1}{2}\left(2x+3y+4z-4\right)=0\Rightarrow y+2=\mathrm{0...}\left(2\right)$
$\therefore$ S.D. = distance of any point say (0, 0, 0) on z-axis from plane (2) $=\frac{2}{\sqrt{{\left(1\right)}^2}}=2$