# Finding equation of plane(s) perpendicular to the line of intersection of two other planes.

Question

Find the Cartesian equation of the plane(s) 2 units from the point A(0, 2, -1) and perpendicular to the line of intersection of the planes given by r = (3, -2, 2) + s(-2, 1, 0) + t(4, -3, 2), s, t elements of Real Numbers and 3x + 3y + 4z - 4 = 0.

My Strategy

I first found the normal of plane 1.

n1 = (2, 4, 2) by crossing the two direction vectors


Then concluded its scalar equation:

1x + 2y + 1z + D = 0
1(3) + 2(-2) + 1(2) + D = 0
x + 2y + z - 1 = 0


By eliminating one variable from both scalar equations and writing both equations in terms of y = s, element of Real Numbers, I found the parametric equations of their line of intersection.

x = -5s
y = s
z = 3s + 1


My Problem

From this point on, I do not know how to fit the remaining plane criteria to solve the rest of the question, relating to point A and the plane being perpendicular to the line of intersection determined above.

Any help solving this is greatly appreciated! Please keep in mind I am a high school student so minimum complexity would be awesome.

Thanks!