What condition must hold among A_i,B_i,C_i,D_i for the four…
What condition must hold among $A_i,B_i,C_i,D_i$ for the four planes $A_ix+B_iy+C_iz+D_i=0$ ($i=1,\dots,4$) to pass through a common point?
Explanation of Method
The condition for four planes to pass through a common point.
Step 1: Necessary condition
If $(x_0,y_0,z_0)$ is common, then $(x_0,y_0,z_0,1)\ne\vec0$ solves the homogeneous $4\times4$ system with matrix $\widetilde M=(A_i\ B_i\ C_i\ D_i)$. Hence
\[\begin{vmatrix}A_1&B_1&C_1&D_1\\A_2&B_2&C_2&D_2\\A_3&B_3&C_3&D_3\\A_4&B_4&C_4&D_4\end{vmatrix}=0\]
Expanding along the 4th row gives $A_4\Delta_x+B_4\Delta_y+C_4\Delta_z+D_4\Delta_{123}=0$ (with the Cramer determinants of the first three planes). This says exactly that the fourth plane passes through the point of the first three.
Step 2: Necessary and sufficient condition
In general (Kronecker–Capelli for the $4\times3$ system): $\operatorname{rank}M=\operatorname{rank}\widetilde M,\qquad M=(A_i\ B_i\ C_i)_{4\times3}$.
In the usual situation, some three of the planes have $\Delta\ne0$, so $\operatorname{rank}M=3$. Then $\det\widetilde M=0$ is necessary and sufficient, and the common point is unique.
Result
\[\boxed{\,\begin{vmatrix}A_1&B_1&C_1&D_1\\A_2&B_2&C_2&D_2\\A_3&B_3&C_3&D_3\\A_4&B_4&C_4&D_4\end{vmatrix}=0\,}\]
\[\boxed{\,\operatorname{rank}M=\operatorname{rank}\widetilde M,\qquad M=(A_i\ B_i\ C_i)_{4\times3}\,}\]
Remark
Correction: the source presents $\det\widetilde M=0$ (equivalently $\operatorname{rank}\widetilde M\le3$) as the general necessary and sufficient condition. Without $\operatorname{rank}M=3$ it is only necessary. For example, the planes $x=0,\ x=1,\ x=2,\ y=0$ have $\det\widetilde M=0$ (the $z$-column is zero), but $\operatorname{rank}M=2\lt \operatorname{rank}\widetilde M=3$, and there is no common point.