Question

Find an equation of the plane.The plane through the point(4, −3, −1)and parallel to the plane 6x − y − z = 6

Answer 1

Both planes have the same normal vector:

$6x-y-z=6\implies\mathbf n=\langle6,-1,-1\rangle$

The plane we want must contain the point (4, -3, -1), so its equation would be

$\mathbf n\cdot\langle x-4,y+3,z+1\rangle=0$

$\implies 6(x-4)-(y+3)-(z+1)=0$

$\implies\boxed{6x-y-z=28}$