Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of the perpendicular distance of origin from L is ?

Let $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ be the required line.
Then $2l+3m-10n=0,4l-3m-2n=0$
So $l:m:n=1:1:\frac{1}{2}$
So the line is $\frac{x-x_1}{1}=\frac{y-y_1}{1}=\frac{z-z_1}{\frac{1}{2}}$.
But i could not solve further.Please help me.

Hint.
I give you a hint that you can apply to find the distance you're looking for.
So you have two lines $$\begin{cases} L_1 \equiv \frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10} \\
L_2 \equiv \frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}\end{cases}$$
For $L_1$, you can find a point $A_1 \in L_1$, for example $A_1=(-2,-6,34)$ and a direction vector of $L_1$, for example $u_1=(2,3,-10)$. You can do the same for $L_2$ to get $A_2$ and $u_2$.
Now the cross product $u = u_1 \times u_2$ is perpendicular to $u_1$ and $u_2$. The plane $P_1$ containing the line $L_1$ and the direction $u$ intersects $L_2$ at $B_2$. Similarly, The plane $P_2$ containing the line $L_2$ and the direction $u$ intersects $L_1$ at $B_1$. The line $B_1 B_2$ is the perpendicular to your orginal two lines.
Finally you can find its distance to the origin.
