What condition must hold among A_i,B_i,C_i,D_i for the…
What condition must hold among $A_i,B_i,C_i,D_i$ for the planes $A_ix+B_iy+C_iz+D_i=0$ ($i=1,2,3$) to belong to a pencil, i.e. to intersect along a single common line?
Explanation of Method
The condition for three planes to belong to one pencil, i.e. to share exactly one common line.
Step 1: Condition
The solution set of the system must be a line: consistent, with one free parameter. Hence
\[\operatorname{rank}\begin{pmatrix}A_1&B_1&C_1\\A_2&B_2&C_2\\A_3&B_3&C_3\end{pmatrix}=\operatorname{rank}\begin{pmatrix}A_1&B_1&C_1&D_1\\A_2&B_2&C_2&D_2\\A_3&B_3&C_3&D_3\end{pmatrix}=2\]
In determinant language this means:
- $\Delta=\det M=0$;
- some $2\times2$ minor of $M$ is non-zero;
- the $3\times3$ minors of $\widetilde M$ that contain the $D$-column vanish.
Step 2: Linear-combination form
Suppose the first two planes are not parallel, i.e. $\vec n_1\nparallel\vec n_2$. Then the condition is that the third equation is a combination of the first two: $(A_3,B_3,C_3,D_3)=\lambda(A_1,B_1,C_1,D_1)+\mu(A_2,B_2,C_2,D_2)$.
Step 3: Overview
| $\operatorname{rank}\widetilde M$ | $\operatorname{rank}M$ | Configuration |
|---|---|---|
| 3 | 3 | unique common point |
| 3 | 2 | no common point (e.g. a triangular prism) |
| 2 | 2 | common line: pencil |
| 2 | 1 | no common point (parallel planes, not all equal) |
| 1 | 1 | all three planes coincide |
Result
\[\boxed{\,\operatorname{rank}\begin{pmatrix}A_1&B_1&C_1\\A_2&B_2&C_2\\A_3&B_3&C_3\end{pmatrix}=\operatorname{rank}\begin{pmatrix}A_1&B_1&C_1&D_1\\A_2&B_2&C_2&D_2\\A_3&B_3&C_3&D_3\end{pmatrix}=2\,}\]
Remark
Corrections:
- The source's “compact condition” $\operatorname{rank}\widetilde M=2$ alone is not sufficient. With $\operatorname{rank}M=1$ it describes parallel planes with no common point, so $\operatorname{rank}M=2$ must also be required.
- In the last row of the source's table, rank 1 means the planes coincide. Parallel distinct planes have $\operatorname{rank}\widetilde M=2$.