To find the equation of a plane

Question

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 $$(x-3)/-2 = (y+3)/8 = (z+2)/-10$$

Answer 1

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.

Answer 2

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.