Example 4 In the arithmetic sequence $\left\{a_{n}\right\}$, $a_{l}=\frac{1}{a}, a_{m}=\frac{1}{b}, a_{n}=\frac{1}{c}$, try to find the value of $(l-m) a b+$ $(m-n) b c+(n-l) c a$.
In the Cartesian coordinate system, for any $n \in \mathbf{N}_{+}$, the points $\left(n, a_{n}\right)$ are collinear, so the points $\left(l, \frac{1}{a}\right)$, $\left(m, \frac{1}{b}\right)$, and $\left(n, \frac{1}{c}\right)$ are collinear. Thus, $\frac{\frac{1}{b}-\frac{1}{a}}{m-l}=\frac{\frac{1}{c}-\frac{1}{b}}{n-m}=\frac{\frac{1}{c}-\frac{1}{a}}{n-l}$. From $\frac{\frac{1}{c}-\frac{1}{a}}{n-l}=$ $\frac{\frac{1}{c}-\frac{1}{b}}{n-m}$, simplifying gives $a c(n-l)=\frac{b c(m-n)(c-a)}{b-c}$, so $(l-m) a b+(m-$ $n) b c+(n-l) c a=\frac{b c(m-n)(b-a)}{c-b}+b c(m-n)-\frac{b c(m-n)(c-a)}{c-b}=$ $b c(m-n) \cdot \frac{b-c}{c-b}+b c(m-n)=-b c(m-n)+b c(m-n)=0$. Therefore, the required value is 0.