x-3/3=y/0=z/-4=$(let) (given)
A(3$+3,0,-4$) is any point on given line
Dc of y axis 0,1,0
Any point on y axis B(0,k,0)
DC of AB 3$+3-0,0-k,-4$-0
For shortest distance
(3$+3)*3+(-k)*0+(-4$)*(-4)=0
So,$=-9/25
Also,
(3$+3)*0+-k*1+(-4$)*0=0
So,k=0
Therefore shortest distance between the lines is distance between
A(48/25,0,-36/25) and B(0,0,0)
=12/5