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

Any two points on the line x+y+2z=0, 2x+3y+4z=0 are (1,-1,0) and (5,-1,0).
Thus its equation may be written as, x-1 = y+1 = z/0
and equation of line z-axis is, x=0, y=0, z=1
Now we know shortest distance between two lines is given by,
D = |(a-c).(b x d)|/|b x d|
Here a=(0,0,0), d=(1,-1,0), b=(1,1,0), c=(1,-1,0)
∴ b x d = | i j k |
| 1 1 0 |
| 1 -1 0 |
= 0i + 0j -2k
Thus D = |(1-1)i + (-1+1)j + (0-0)k|.|-2k|/|-2k| = 0/2 = 2