$7 \cdot 69$ Let set $M$ consist of all points in the plane except for 3 distinct points $A, B, C$. Find the minimum number of convex sets such that their union equals $M$.

The translation is as follows:

$7 \cdot 69$ Let set $M$ consist of all points in the plane except for 3 distinct points $A, B, C$. Find the minimum number of convex sets such that their union equals $M$.
[Solution] First, let points $A, B, C$ be collinear on line $l$, then they divide the line $l$ into 4 intervals, and points in different intervals cannot belong to the same convex set, so there must be at least 4 convex sets. On the other hand, when taking $l$ as the $x$-axis and the coordinates of $A, B, C$ as $\left(x_{1}, 0\right),\left(x_{2}, 0\right),\left(x_{3}, 0\right)\left(x_{1}\right.$ $0\} \cup\left\{(x, 0) \mid xx_{3}\right\}, \\
M_{3}=\left\{(x, 0) \mid x_{1}<x<x_{2}\right\}, M_{4}=\left\{(x, 0) \mid x_{2}<x<x_{3}\right\},
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It is easy to verify that they are all convex sets and their union is $M$. Therefore, the minimum number of convex sets required in this case is 4.

Now, assume points $A, B, C$ are not collinear, then points $B$ and $C$ divide the line $BC$ into 3 intervals, and points in different intervals should belong to different convex sets. Therefore, there must be at least 3 convex sets. On the other hand, when dividing $M$ into 3 convex sets $M_{1}$, $M_{2}$, $M_{3}$ as shown in the right figure, the requirement is met, where $M_{1}$ includes the open segment $A \dot{C}, BC$, $M_{2}$ includes all boundary points except the segment $AC$, and $M_{3}$ is the lower half-plane containing the ray emitted from point $B$ to the left (excluding point $B$). It is evident that the minimum number of convex sets required in this case is 3.