Example 8 Solve the system of equations in the set of real numbers
$$\left\{\begin{array}{l}
2 x+3 y+z=13 \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82
\end{array}\right.$$
By adding the left and right sides of the two equations and transforming, we get
$$(2 x)^{2}+(3 y+3)^{2}+(z+2)^{2}=108$$

From the equation $2 x+3 y+z=13$ transformed, we get
$$2 x+(3 y+3)+(z+2)=18,$$

Thus, by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
18^{2} & =[1 \times(2 x)+1 \times(3 y+3)+1 \times(z+2)]^{2} \\
& \leqslant\left(1^{2}+1^{2}+1^{2}\right) \times\left[(2 x)^{2}+(3 y+3)^{2}+(z+2)^{2}\right] \\
& =3 \times 108=324=18^{2}
\end{aligned}$$

The equality holds if and only if: $\frac{2 x}{1}=\frac{3 y+3}{1}=\frac{z+2}{1}$.
Noting that $2 x+(3 y+3)+(z+2)=18$,
$$\therefore 2 x=3 y+3=z+2=6 \text {, }$$

Solving, we get $x=3, y=1, z=4$.
Therefore, the solution to the original system of equations is $\left\{\begin{array}{l}x=3, \\ y=1, \\ z=4 .\end{array}\right.$