What condition must hold among A_i,B_i,C_i,D_i for the three…
What condition must hold among $A_i,B_i,C_i,D_i$ for the three planes $A_ix+B_iy+C_iz+D_i=0$ ($i=1,2,3$) to pass through a common point?
Explanation of Method
The condition on the coefficients for three planes to have a common point.
Step 1: Condition (Kronecker–Capelli)
Let $M=(A_i\ B_i\ C_i)$ be the $3\times3$ coefficient matrix and $\widetilde M=(A_i\ B_i\ C_i\ D_i)$ the $3\times4$ augmented matrix. A common point exists iff
\[\operatorname{rank}M=\operatorname{rank}\widetilde M\]
Using $D_i$ or $-D_i$ in the last column does not change the rank.
Step 2: Cases
| Ranks | Common points |
|---|---|
| $\Delta=\det M\ne0$ (rank 3) | exactly one point, for any $D_i$ |
| $\operatorname{rank}M=\operatorname{rank}\widetilde M=2$ | a common line (pencil) |
| $\operatorname{rank}M=\operatorname{rank}\widetilde M=1$ | the three planes coincide |
| $\operatorname{rank}M\lt \operatorname{rank}\widetilde M$ | none |
When $\Delta\ne0$, Cramer's rule gives the point: \[x_0=\frac{\Delta_x}{\Delta},\quad y_0=\frac{\Delta_y}{\Delta},\quad z_0=\frac{\Delta_z}{\Delta},\] where $\Delta_x,\Delta_y,\Delta_z$ are obtained by replacing a column of $\Delta$ with $(-D_1,-D_2,-D_3)^T$.
When $\Delta=0$ and $\operatorname{rank}M=2$, consistency requires the $3\times3$ minors of $\widetilde M$ containing the $D$-column to vanish. In fact one of them suffices: any minor formed with two independent columns of $M$.
Result
\[\boxed{\,\operatorname{rank}M=\operatorname{rank}\widetilde M\,}\]