distance between 2 lines in 3d 
Calculate the distance between the lines
  $$L_1:x=4+2t,y=3+2t,z=3+6t$$
  $$L_2: x=-2+3s ,y=3+4s ,z=5+9s$$

I tried subtraction $L_1$ from $L_2$ then multiplying the resting vector by the $t$'s and $s$'s original values and trying to find value for $t$ to or $s,$ but I found $t=\frac{19}{12} s$ and I don't know how to keep solving this.

$ \hat a = 4\hat i + 3\hat j+3\hat k$
$ \hat b = -2\hat i + 3\hat j+5\hat k$
$\hat t =2\hat i + 2\hat j+6\hat k$
$\hat s = 3\hat i + 4\hat j+9\hat k$
$L1:  \hat a+t\hat t$
$L2:  \hat b+s\hat s$
So the distance $$d = \dfrac{\left|(\hat a - \hat b).(\hat t \times \hat s)\right|}{|\hat t \times \hat s|}$$
$\hat t \times \hat s = -6\hat i + 2\hat k$
$ \hat a - \hat b =  6\hat i - 2\hat k$
$\left|(\hat a - \hat b).(\hat t \times \hat s)\right| = 40$
$|\hat t \times \hat s| = \sqrt{40}$
Thus the distance $= \frac{40}{\sqrt{40}}$
$d = \sqrt{40} = 2\sqrt{10}$
