Example 7 Given $a+b+c=1$,
$$
\frac{1}{a+1}+\frac{1}{b+3}+\frac{1}{c+5}=0 \text {. }
$$

Find the value of $(a+1)^{2}+(b+3)^{2}+(c+5)^{2}$.
【Analysis】The denominator in equation (1) is a bit complicated, so consider using the substitution method.
Solution Let $x=a+1, y=b+3, z=c+5$.
Then the given equations can be transformed into
$$
\begin{array}{l}
x+y+z=10, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \Rightarrow x y+y z+z x=0 .
\end{array}
$$

From formula (1) we get
$$
\begin{array}{l}
(a+1)^{2}+(b+3)^{2}+(c+5)^{2} \\
=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
=100
\end{array}
$$