The equation of any plane passing through given line is
$\left(x + y+2z - 3\right) + \lambda \left(2x + 3 y + 4z-4\right) = 0$
$\Rightarrow \left(1+2\lambda\right)x+\left(x+3\lambda\right) y+\left(2+4\lambda\right)z-\left(3+4\lambda\right)=0$
If this plane is parallel to z-axis then normal to the plane will be perpendicular to z-axis.
$\therefore \left(1+2\lambda\right)\left(0\right)+\left(1+3\lambda\right)\left(0\right)\left(2+4\lambda \right)\left(1\right)=0$
$\lambda=-\frac{1}{2}$
Thus, Required plane is
$\left(x+y+2z-3\right)-\frac{1}{2}\left(2x +3y+4z-4\right)=0$
$\Rightarrow y+2=0$
$\therefore S.D=\frac{2}{\sqrt{\left(1\right)^{2}}}=2$