Finding three planes which intersect a point with lines

[horses]

Homework Statement

The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). How can I find three planes which also intersect in the point (1; 1; 1) such that each plane contains one and only one of the three lines?

Homework Equations

plane aix + biy + ciz = di

The Attempt at a Solution

I get 9 equations:

Sharing equations with the lines:

a1(1 - t) + b1(1 + 2*t) + c1(1 + t) = d1
a2(u) + b2(2*u - 1) + c2(3*u - 2) = d2
a3(v - 1) + b3(2*v - 3) + c3(3 - v) = d3

Intersection at (1,1,1):

a1 + b1 + c1 = d1
a2 + b2 + c2 = d2
a3 + b3 + c3 = d3

Dot product of plane normals and line vectors = 0 since perpendicular:

<a1; b1; c1> dot <-1; 2; 1> = -a1 + 2*b1 + c1 = 0
<a2; b2; c2> dot <1; 2; 3> = a2 + 2*b2 + 3*c2 = 0
<a3; b3; c3> dot <1; 2; -1> = a3 + 2*b3 - c3 = 0

I know how to find the intersection of 3 planes using matrices/row reduction, and I know some relationships between lines and planes. However, I seem to come up with 12 unknowns and 9 equations for this problem. I know the vectors for the lines must be perpendicular to the normals of the planes, thus the dot product between the two should be 0. I also know that the planes pass through the point (1,1,1) and the x,y,z coordinates for the parameters given in the line equations. What information am I missing? Maybe there are multiple solutions. If so, how can these planes be described with only a line and one point? Another thought was to convert the planes to parametric form, but to describe a plane with parameters normally I would have 2 vectors and one point, but here I only have one vector and one point.

 

[horses]

this can be closed as I slept on it and figured it out...

once you know the vectors you can find the planes since there are multiple answers you can choose values of a,b,c as long as they satisfy the perpendicular equation and going through the point 1,1,1.