3D geometry-skew lines, distance of a point from a line A straight line L  intersects perpendicularly both the line:
$$\frac{(x+2)}{2} = \frac{(y+6)}{3}=\frac{(z-34)}{-10} $$
and the line:
$$\frac{(x+6)}{4}=\frac{(y-7)}{-3}=\frac{(z-7)}{-2}$$
Then the square of the perpendicular distance of origin from L is
I could find the shortest distance between the skew lines but I can't find the equation of line and also the position of origin. How is the origin related to this system, I have no idea. Please help. 

$$L_1: \frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}=a$$
On $L_1$, a general point is $\vec p_1=(2a-2,3b-6,-10a+34)$ 
$$L_2: \frac{x+6}{4}=\frac{x-7}{-3}=\frac{x-7}{-2}=b$$
A generat point on $L_2$ is $\vec p_2\ (4b-6,-3b+7,-2b+7)$
Let the line $L_3$ which is  $P_1P_2$, intersects both $L_1$ and $L_2$ orthogonally, 
then $$(\vec p_1- \vec P_2).\vec L_1 =0 = (\vec P_1- \vec P_2). \vec L_2$$ 
We get $$113a-19b=301,~~ 19a-29b=-1 \implies a=3, b=2$$
So the point $\vec P_1=(4,3,4), \vec P_2=(2,1,3)$
We get line $L_3$ which joins $P_1,P_2$ as
$$L_3: \frac{x-4}{2}=\frac{y-3}{2}=\frac{z-4}{1}=c$$
with a gereral point $\vec p=(2c+4,2c+3, c+4)$ on it.
Then $\vec {OP}$ is perpendicular to $L_3$, so
we get $2(2c+4)+2(2c+3)+1(c+4)=0 \implies c=-2$ giving us the foot of perpendicular as $(0,-1,2)$, its distance from orogin is $\sqrt{5}$ which is the required distance.
