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.
 A: The distance between the lines
$L_1:x=4+2t,y=3+2t,z=3+6t$ and $L_2: x=-2+3s ,y=3+4s ,z=5+9s$ is
$d=2 \sqrt{10}$
Indeed consider the function which gives the distance between a generic point of the first line and a generic point of the second line
$f(t,s)=\sqrt{(2 t-4 s)^2+(-3 s+2 t+6)^2+(-9 s+6 t-2)^2}$
and set to zero the partial derivatives $\partial f_t=0,\partial f_s=0$
$\left\{
\begin{array}{l}
 4 (2 t-4 s)+4 (-3 s+2 t+6)+12 (-9 s+6 t-2)=0 \\
 -8 (2 t-4 s)-6 (-3 s+2 t+6)+18 (-9 s+6 t-2)=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
 17 s-11 t=0 \\
 53 s-34 t=0 \\
\end{array}
\right.$
which has one solution
$t= 0,\;s=0$
we have
$f(0,0)=2 \sqrt{10}$
Hope it helps
A: A general solution is :
 take a point on $l_1 :A=(4+2t,3+2t,3+6t)$
take a point on $L_2: B=(-2+3s ,3+4s ,9s)$  
find vector $\overline{AB}=B-A=\\(-2+3s ,3+4s ,9s)-(4+2t,3+2t,3+6t)$   
then solve the system of equation $$\begin{cases}\overline{AB}.V_{l_1}=0\\\overline{AB}.V_{l_2}=0\end{cases}\\
\begin{cases}\overline{AB}.(2,2,6)=0\\\overline{AB}.(3,4,9)=0\end{cases}$$ you will have two unknown and two linear equation ,here . $|\overline{AB}|$ is the minimum distance between two lines.
A: $ \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 $= \frac{40}{\sqrt{40}}$
$d = \sqrt{40} = 2\sqrt{10}$
