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This question already has an answer here:

How to find an equation of a plane when 2 lines lying on the plane are given ?
Q)Find the equation of plane which contains the lines $$(x-4)/1 = (y-3)/-4 = (z-2)/5$$ and $$(x-3)/-2 = (y+3)/8 = (z+2)/-10$$

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marked as duplicate by amd, Lord Shark the Unknown, Brahadeesh, GNUSupporter 8964民主女神 地下教會, Shailesh Oct 8 '18 at 9:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Oct 7 '18 at 8:26
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Normal to your plane is common perpendicular to both lines.

Direction vectors are $ (3,-4,5)$ and $(-2,8,-10)$

The cross product of these two vectors is the normal to the plane.

Find one point on the plane and use the normal vector to write the equation of the plane.

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Given the symmetric equations (see here) for the lines, we can deduce two points and two vectors that lie on the plane. Let $r$ be the first line and $s$ the second. Then, in vector form, $$ r:\quad (4,3,2)+\lambda(3,-4,5)\\ s:\quad (3,-3,2) + \mu(-2,8,-10)\,. $$ Since the vectors $(3,-4,5)$ and $(-2,8,-10)$ lie on the plane and are linearly independent (i.e. they're not multiples of one another), every point on the plane can be "found" through a linear combination of these two vectors. Try to continue from this point.

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