Example 2 Let $A, B, C$ be three non-collinear lattice points on a plane, and the side lengths of $\triangle ABC$ are all positive integers. Find the minimum value of $AB$ and the minimum perimeter.
【Analysis】If $A B=1$, we might as well set $A(0,0), B(1,0)$. Then $|A C-B C|<A B=1$, which can only be $A C=B C$.
Thus, point $C$ lies on the perpendicular bisector of $A B$.
Therefore, the x-coordinate of point $C$ is $\frac{1}{2}$, meaning $C$ cannot be a lattice point.
Hence, $A B$ cannot be 1.
If $A B=2$, we might as well set $A(0,0), B(2,0)$.
Then $|A C-B C|<2$, which can only be $|A C-B C|=0,1$.
If $A C=B C \Rightarrow x_{c}-1 \Rightarrow A C^{2}=1+y_{c}^{2}$, thus, the difference of two perfect squares cannot be 1.
If $B C-A C=1$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
A C^{2}=x_{c}^{2}+y_{c}^{2}=k^{2}, \\
B C^{2}=\left(x_{C}-2\right)^{2}+y_{C}^{2}=(k+1)^{2}
\end{array}\right. \\
\Rightarrow 2 k+1=4-4 x_{C} .
\end{array}
$$

The parity of both sides of the above equation is different, so the equation does not hold, i.e., $B C-A C$ is not equal to 1.
Therefore, $A B$ cannot be 2.
If $A B=3$, we might as well set $A(0,0), B(3,0)$. Then $C(3,4)$ satisfies the conditions of the problem.
Therefore, the minimum value of $A B$ is 3.
Noting that the three vertices of $\triangle A B C$ are lattice points, using the area formula of analytic geometry, we know
$$
2 S_{\triangle A B C}=\left(\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right)
$$

The absolute value is a positive integer.
Also, the three sides of $\triangle A B C$ are all integers, with side lengths greater than or equal to 3, thus, using Heron's formula, the side lengths of 3, 3, 3; 3, 3, 4; 3, 4, 4; and 3, 3, 5 are all impossible (the area is irrational); 3, 4, 5 satisfies the conditions.
Therefore, the minimum perimeter is 12.