Example 1. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{a+\lambda}+\frac{y}{b+\lambda}+\frac{z}{c+\lambda}=1, \\
\frac{x}{a+\mu}+\frac{y}{b+\mu}+\frac{z}{c+\mu}=1, \\
\frac{x}{a+\gamma}+\frac{y}{b+\gamma}+\frac{z}{c+\gamma}=1 .
\end{array}\right.
$$
Given the system of equations, we know that $\lambda, \mu, \gamma$ are the roots of the equation in the variable $\theta$:
$$
\frac{x}{\theta+a}+\frac{y}{\theta+b}+\frac{z}{\theta+c}=1
$$

Let $\theta=t-a$, then (1) transforms into a cubic equation in the variable $t$:
$$
t^{3}+\cdots-(a-b)(a-c) x=0. \quad \text { (*) }
$$

Considering that $\theta=\lambda, \mu, \gamma$ are the three roots of equation (1) in the variable $\theta$, the roots of equation (*) in $t$ are $t=a+\theta$, which are $a+\lambda, a+\mu, a+\gamma$. Therefore, by the generalized Vieta's theorem, the product of the roots of equation (*) is
$$
\begin{aligned}
& (a+\lambda)(a+\mu)(a+\gamma) \\
= & (a-b)(a-c) x,
\end{aligned}
$$

Solving for $x$ gives $x=\frac{(a+\lambda)(a+\mu)(a+\gamma)}{(a-b)(a-c)}$,
If we let $\theta=t-b, \theta=t-c$, and repeat the above solution process, we can find
$$
\begin{array}{l}
y=\frac{(b+\lambda)(b+\mu)(b+\gamma)}{(b-a)(b-c)}, \\
z=\frac{(c+\lambda)(c+\mu)(c+\gamma)}{(c-a)(c-b)}
\end{array}
$$