Example 5 Find all solutions to $15 x_{1}+10 x_{2}+6 x_{3}=61$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
To solve for the coefficient of $x_{3}$ with the smallest absolute value, we transform the original equation into
$$\begin{aligned}
x_{3} & =\frac{1}{6}\left(-15 x_{1}-10 x_{2}+61\right) \\
& =-2 x_{1}-2 x_{2}+10+\frac{1}{6}\left(-3 x_{1}+2 x_{2}+1\right)
\end{aligned}$$

The original indeterminate equation is equivalent to the following indeterminate equation
$$x_{4}=\frac{1}{6}\left(-3 x_{1}+2 x_{2}+1\right) \in \boldsymbol{Z} .$$

Using a similar method, we get
$$\begin{aligned}
x_{2} & =\frac{1}{2}\left(6 x_{4}+3 x_{1}-1\right) \\
& =3 x_{4}+x_{1}+\frac{1}{2}\left(x_{1}-1\right), \\
x_{5} & =\frac{1}{2}\left(x_{1}-1\right) \in \boldsymbol{Z} .
\end{aligned}$$

Solving this, we get
$$x_{1}=1+2 x_{5}, \quad x_{5}=0, \pm 1, \pm 2, \cdots$$

By back-substitution, we solve sequentially
$$\begin{array}{l}
x_{2}=3 x_{4}+x_{1}+x_{5}=1+3 x_{4}+3 x_{5} \\
x_{4}, x_{5}=0, \pm 1, \pm 2, \cdots \\
x_{3}=-2 x_{1}-2 x_{2}+10+x_{4} \\
\quad=6-5 x_{4}-10 x_{5} \\
x_{4}, x_{5}=0, \pm 1, \pm 2, \cdots
\end{array}$$

This gives the general solution to the original indeterminate equation, which contains two parameters $x_{4}, x_{5}$.