The equation of the line l can also be written:
<0,-2,2>+λ<2,3,-1> where <2,3,-1> is the direction vector, λ is a scalar quantity determining points on the line. A line cannot define a plane because there are many planes (an infinitude) containing l. We need three non-collinear points to define a plane. To find two points on l, use λ=0 and λ=1: A(0,-2,2) and B(2,1,1).
The equation of a line m passing through P(2,0,-1) and parallel to l is:
<2,0,-1>+μ<2,3,-1>, where μ is a scalar variable, like λ.
When μ=1 we get another point on the same line: Q(4,3,-2). The lines l and m are parallel so all 4 points can lie on the same plane. The normal of the plane is also normal to all vectors lying in the plane.
So, we can define a plane containing A, B and P.
If O(0,0,0) is the origin of the frame of reference for points on l, then the following vector equations apply:
OA=<0,-2,2>=2i+0j-k, where i, j, k are unit vectors in the x, y, z directions.
OB=<2,1,1>=2i+j+k. Since OA+AB=OB, AB=OB-OA=<2,3,-1> (which is, of course, the direction vector of l).
OP=<2,0,-1>, OQ=<4,3,-2>. OA+AP=OP, AP=OP-OA=<2,2,-3>.
OB+BP=OP, BP=OP-OB=<0,-1,-2>.
If the normal vector n to each is <a,b,c>, then:
<2,3,-1>.n=<2,2,-3>.n=<0,-1,-2>.n=0 (dot products).
Therefore 2a+3b-c=2a+2b-3c=-b-2c=0, from which b=-2c, and 2a=7c.
If c=2, then b=-4 and a=7, n=<7,-4,2>. The equation of the plane, which has the same normal as all vectors on the plane, is derived from the normal:
7x-4y+2z=k. To determine k, simply plug in the coordinates of any point in the plane (that is, any of the points A, B, P, Q).
Let's use A(0,-2,2): k=8+4=12, making the equation of the plane 7x-4y+2z=12.
Now check the other points for consistency:
B(2,1,1): k=14-4+2=12; P(2,0,-1): k=14-0-2=12; Q(4,3,-2): k=28-12-4=12. So all points lie on the plane.
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