Hi, I'm having trouble answering this question.

Question

The lines l1 and l2 are defined as
l1: (x-1)/3 = (y-5)/2 = (z-12)/-2
l2: (x-1)/8 = (y-5)/11 = (z-12)/6

The plane j contains both l1 and l2. Find the Cartesian equation of j.

Answer 1

Take the cross product and the normal to both lines is <2,-2,1>

The plane is then just

2(x-1)-2(y-5)+1(z-12) = 0
2x-2y+z = 4