Example 6 Find the solution to $15 x_{1}+10 x_{2}+6 x_{3}=61$.

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Solve using the method of Theorem 3. $a_{1}=15, a_{2}=10, a_{3}=6$, so $g_{1}=$ $a_{1}=15, g_{2}=\left(a_{1}, a_{2}\right)=(15,10)=5, g_{3}=\left(g_{2}, a_{3}\right)=(5,6)=1$. Therefore, this indeterminate equation is equivalent to a system of indeterminate equations with 4 variables and two equations:
$$\left\{\begin{array}{l}
5 y_{2}+6 x_{3}=61, \\
15 x_{1}+10 x_{2}=5 y_{2}
\end{array}\right.$$
The general solution of $15 x_{1}+10 x_{2}=5 y_{2}$ is
$$x_{1}=y_{2}+2 t_{1}, \quad x_{2}=-y_{2}-3 t_{1}, \quad t_{1}=0, \pm 1, \cdots$$
The general solution of $5 y_{2}+6 x_{3}=61$ is
$$y_{2}=5+6 t_{2}, \quad x_{3}=6-5 t_{2}, \quad t_{2}=0, \pm 1, \cdots$$

Eliminating $y_{2}$ yields the general solution of the original indeterminate equation:
$$\begin{array}{c}
x_{1}=5+2 t_{1}+6 t_{2}, \quad x_{2}=-5-3 t_{1}-6 t_{2}, \quad x_{3}=6-5 t_{2}, \\
t_{1}, t_{2}=0, \pm 1, \cdots .
\end{array}$$