I want to find 3 planes that each contain one and only one line from a set 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.
 A: A little background:
The equation of a plane can be expressed as $\langle n, x \rangle = \pi$, where $n$ is a normal to the plane. If the plane contains the line $t \mapsto x_0 + t d$, then it is fairly straightforward to show that the plane contains the line iff $\langle n, x_0 \rangle = \pi$, and $\langle n, d \rangle = 0$.
Now let $d_1 = (-1,2-1)^T$, $d_2 = (1,2,3)^T$, $d_3 = (1,2 ,-1)^T$, and $e = (1,1,1)^T$. Then the three lines are $t \mapsto e + t d_i$, $i = 1,2,3$.
The problem is to find three normals $n_i$ such that $\langle n_i, d_j \rangle = 0$ iff $i = j$. If we can find these normals, then we can set $\pi_i = \langle n_i, e \rangle$, and the three planes will be given by $\langle n_i, x \rangle = \pi_i$.
Let $N = \begin{bmatrix}n_1 & n_2 & n_3\end{bmatrix}$, and $D = \begin{bmatrix}d_1 & d_2 & d_3\end{bmatrix}$. Then we want to find a matrix $N$ such that the following equation holds, where $*$ represents any non-zero number):
$$N^T D = \Delta =\begin{bmatrix}0 & * & * \\ * & 0 & * \\ * & * & 0\end{bmatrix}.$$
Since $D$ is invertible (check!), the solution is given by $N = D^{-T} \Delta^{T}$, with the constants $\pi_i$ given by $(\pi_1, \pi_2, \pi_3)^T = N^T e$.
For example, choose $\Delta =\begin{bmatrix}0 & 4 & 4 \\ 4 & 0 & 4 \\ 4 & 4 & 0\end{bmatrix}$. Then $N = \begin{bmatrix}2 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & -1 & 1\end{bmatrix}$, and $(\pi_1, \pi_2, \pi_3)  = (3, 0, 1)$.
A: Avoid writing down so many equations to be solved, but produce these planes in a forward motion: Your three lines $\ell_i$ $\ (1\leq i\leq3)$ can be given as
$$\ell_i:\quad t\mapsto a+ t\ p_i \qquad(-\infty<t<\infty)$$
with $a=(1,1,1)$, $p_1=(-1,2,1)$, $p_2=(1,2,3)$, $p_3=(1,2,-1)$, where now all three lines pass the point $a$ at time $t=0$.
The plane $\pi_1$ with parametric representation
$$\pi_1:\quad (u,v)\mapsto a + u p_1 +v{p_2+p_3\over 2}\qquad\bigl((u,v)\in{\mathbb R}^2\bigr)$$
contains the line $\ell_1$ and is transversal to $\ell_2$ and $\ell_3$. (In order to visualize this imagine that the $p_i$ are the three standard basis vectors.) Define $\pi_2$ and $\pi_3$ similarly.
It is easy to convert the parametric representations of the $\pi_i$ into equations, if desired.
