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Can someone help me solve the following two questions:

1) Find the distance between the lines: $$ L_1: \frac{x-1}{2} = \frac{y+3}{1} = \frac{z}{-1}$$ and $$\displaystyle L_2 : \frac{x+2}{-2}=\frac{y+5}{3} = \frac{z-1}{-5} $$ I've tried taking an arbitrary point on $L_1$ and build the line passing through it and prependcular to $L_1$ , but couldn't figure out how to do it.

2) Find the equation of the plane passing through $(2,-1,5)$ and perpendicular to the planes $3x-2y+z+7=0$ and $5x-4y+3z+1=0$.

Thanks in advance !

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    For 1: you can use the formula [here](http://en.wikipedia.org/wiki/Skew_lines#Distance_between_two_skew_lines), after a few suitable conversions...2012-05-07
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    For 2: the line of intersection of the planes $P_1:\ 3x-2y+z+7=0$ and $P_2:\ 5x-4y+3z+1=0$ is normal to the plane, $P_3$, that you are trying to find. A direction vector of this line is your normal vector to the plane $P_3$. To get this vector, you may take the cross product of the normal vectors of $P_1$ and $P_2$.2012-05-07

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