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Let $ L_1 $ be the line through the points $ (1, 2, 6) $ and $ (2, 4, 8) $. Let $ L_2 $ be the line of intersection of the planes $ P_1 $ and $ P_2 $, where $ P_1 $ is the plane $ x - y + 2z + 1 = 0 $ and $ P_2 $ is the plane through the points $ (3, 2, -1) $, $ (0, 0, 1) $, and $ (1, 2, 1) $. Calculate the distance between $ L_1 $ and $ L_2 $.

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Yeah. In the question they're saying that let everyone realign that passes through two points 1 to 6 and two for it. And it will be the line of intersection of the planes P one which is x minus Y plus two. Z plus one equal to zero. And another plane having the coordinates three to minus +1001 and 1 to 1. Therefore they have to calculate the distance between line L. one and L. two. Mhm. So for the line L one The direction vector is equal to 1- two. and a normal victor for the plane P one is is equal to 1 -1 and two. Mhm. And the victim from the point 001- 3: -1. For the plane P two From .0012 3 to -1 is equal to 3: -2. Yeah. Yeah. So this is Parlin Too Plain P two. Yeah. And as in the As is the better form from 001 two, 1 to 1. That is 1- zero. Therefore A normal vector for P two is equal to three. 2 -2. Cross 120. Therefore this is equal to 4 -24. Therefore we can simplify this vector for plain P two as prime and playing this with half. So it is equal to two minus one and two. And a direction vector for the line L two. Mhm. Direction. Yeah. Yeah. And direction vector for line L two of intersection of planes B one and P two is V two vector is equal to anyone victor crossed into vectors. So the direction vector for the line of intersection for the line and two of the intersection of the plains is V. Two vector is equal to anyone vector. Cross into vector is equal to 1 -1. 2. Cross 2 -12. This is equal to 0- one. Now we noticed that the point three 2 -1 lies on border prints. So it is, it also lies on L two. The lines are skewed, L one and L two. R Schiewe. So we can view them as lying in two parallel planes. A common normal victims of the plane is N. Is equal to be one trust me to that is equal to -2 -12. And line L one passes through the point 126. So 1, 26 lies on one of the planes and 3 to -1 is a point on L. Do. Therefore this is a point on other plane. Also equations of the Plains then are Mhm Prince -2, X -Y Plus two is equal to eight and -2. X -Y plus two. Said Yeah Equal to -10. Therefore distance between the lines is B is equal to Modular value of -8 minus 10, divided by route under four plus one plus four. That is equal to 18 by three. That is equal to six. So the formula for B is equal to model the value of the one minus the two, divided by route. Under a squared plus B squared plus C square. Where the one is equal to -8, the two is equal to then, and if equal to to be equal to Equal to -2 equal to -1 and C is equal to two. Yeah, so hence the distance between the two lines, L one and L two is equal to six.