Example 2 Find three real numbers $x, y, z$ such that they simultaneously satisfy the following equations:
$$
\begin{array}{l}
2 x+3 y+z=13, \\
4 x^{2}+9 y^{2}+z^{2}-2 x+15 y+3 z=82 .
\end{array}
$$
(1992, Friendship Cup International Mathematics Competition)
Solution: From $(1)$, we get $2 x+3 y=2 \cdot \frac{13-z}{2}$. Then $2 x, \frac{13-z}{2}, 3 y$ form an arithmetic sequence. Therefore, let
$$
2 x=\frac{13-z}{2}+d, 3 y=\frac{13-z}{2}-d .
$$

Substitute into (2), and simplify and complete the square, to get
$3(z-4)^{2}+4\left(d-\frac{3}{2}\right)^{2}=0$.
$\therefore z=4, d=\frac{3}{2}$.
Substitute into $2 x=\frac{13-z}{2}+d$ and $3 y=\frac{13-z}{2}-d$, and solve to get $x=3, y=1$.
Therefore, the three real numbers are $x=3, y=1, z=4$.