Example 9 Let $x, y, z$ all be real numbers. Solve the system of equations
$$
\left\{\begin{array}{l}
2 x+3 y+z=13 \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82
\end{array}\right.
$$
Solve: From $2 x+3 y+z=13$, we get
$$
2 x+3 y=13-z \text {. }
$$

Let $2 x=\frac{13-z}{2}+t, 3 y=\frac{13-z}{2}-t$.
Substituting into equation (1) and simplifying, we get
$$
3(z-4)^{2}+4\left(t-\frac{3}{2}\right)^{2}=0 \text {. }
$$

Therefore, $z=4, t=\frac{3}{2}$.
Thus, $x=3, y=1$.
Hence, the solution to the original system of equations is
$$
(x, y, z)=(3,1,4) \text {. }
$$