13. Real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=\lambda(\lambda>0)$, try to find
$$
f=\min \left\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\right\}
$$

the maximum value.
【Analysis】By the symmetry of $\mathrm{i}$, without loss of generality, assume $a \geq b \geq c$, thus:
$$
a-b>a-c>0,
$$

Therefore, we have:
$$
\begin{aligned}
f & =\min \left\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\right\}=\min \left\{(a-b)^{2},(b-c)^{2}\right\} \\
& \leq(a-b)(b-c) \\
& \leq\left[\frac{(a-b)+(b-c)}{2}\right]^{2} \\
& =\frac{(a-c)^{2}}{4} \\
& \leq \frac{\lambda}{2} .
\end{aligned}
$$

Equality holds if and only if $b=0, a=-c=\frac{\sqrt{2 \lambda}}{2}$.