The three lines intersect in the point $(1, 1, 1)$: $(1 - t, 1 + 2t, 1 + t)$, $(u, 2u - 1, 3u - 2)$, and $(v - 1, 2v - 3, 3 - v)$. 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?
Using the equation for a plane $$a_i x + b_i y + c_i z = d_i,$$ I get $9$ equations.
Sharing equations with the lines:
$$a_1(1 - t) + b_1(1 + 2t) + c_1(1 + t) = d_1,$$ $$a_2(u) + b_2(2u - 1) + c_2(3u - 2) = d_2,$$ $$a_3(v - 1) + b_3(2v - 3) + c_3(3 - v) = d_3.$$
Intersection at $(1,1,1)$: $$a_1 + b_1 + c_1 = d_1,$$ $$a_2 + b_2 + c_2 = d_2,$$ $$a_3 + b_3 + c_3 = d_3.$$
Dot product of plane normals and line vectors is $0$ since perpendicular: $$\langle a_1, b_1, c_1 \rangle \cdot \langle -1, 2, 1 \rangle = -a_1 + 2b_1 + c_1 = 0,$$ $$\langle a_2, b_2, c_2\rangle \cdot \langle 1, 2, 3\rangle = a_2 + 2b_2 + 3c_2 = 0,$$ $$\langle a_3, b_3, c_3 \rangle \cdot \langle 1, 2, -1 \rangle = a_3 + 2b_3 - c_3 = 0.$$
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.