Example 7 Find the real solutions of the system of equations:
$$
\left\{\begin{array}{l}
(x-2)^{2}+\left(y+\frac{3}{2}\right)^{2}+(z-6)^{2}=64, \\
(x+2)^{2}+\left(y-\frac{3}{2}\right)^{2}+(z+6)^{2}=25 .
\end{array}\right.
$$
Solving: By adding and subtracting the two given equations, we get
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=\left(\frac{3}{2}\right)^{2}, \\
-8 x+6 y-24 z=39 .
\end{array}\right.
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
39=-8 x+6 y-24 z \\
\leqslant \sqrt{(-8)^{2}+6^{2}+(-24)^{2}} \cdot \sqrt{x^{2}+y^{2}+z^{2}} \\
=\sqrt{676} \times \sqrt{\frac{9}{4}}=39 .
\end{array}
$$

Furthermore, by the condition for equality in the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
\frac{x}{-8}=\frac{y}{6}=\frac{z}{-24}=\frac{-8 x+6 y-24 z}{64+36+576} \\
=\frac{39}{676}=\frac{3}{52} \text {. } \\
\text { Thus, }\left\{\begin{array} { l } 
{ \frac { x } { - 8 } = \frac { 3 } { 5 2 } , } \\
{ \frac { y } { 6 } = \frac { 3 } { 5 2 } , } \\
{ \frac { z } { - 2 4 } = \frac { 3 } { 5 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=-\frac{6}{13}, \\
y=\frac{9}{26}, \\
z=-\frac{18}{13} .
\end{array}\right.\right. \\
\end{array}
$$

Therefore, the real solution to the system of equations is
$$
(x, y, z)=\left(-\frac{6}{13}, \frac{9}{26},-\frac{18}{13}\right) .
$$