Example 13 Determine the largest real number $z$, such that $x+y+z=5, xy+yz+zx=3$, and $x, y$ are also real numbers.
(7th Canadian Intermediate Mathematics Competition)
Solution: $\because x+y=2 \cdot \frac{5-z}{2}$,
$\therefore x, \frac{5-z}{2}, y$ form an arithmetic sequence.
Thus, let $x=\frac{5-z}{2}+d, y=\frac{5-z}{2}-d$. Substituting into $x y+y z+z x=3$ and simplifying, we get
$$
\begin{array}{l}
3 z^{2}-10 z-13=-d^{2} . \\
\therefore 3 z^{2}-10 z-13 \leqslant 0 .
\end{array}
$$

Solving, we get $-1 \leqslant z \leqslant \frac{13}{3}$.
Furthermore, when $x=y=\frac{1}{3}$, $z=\frac{13}{3}$.
$\therefore$ The maximum value of the real number $z_{\max }=\frac{13}{3}$.