13. (20 points) Given real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=\lambda(\lambda>0)$.
Find $f=\min \left\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\right\}$'s maximum value.
13. Let $a \leqslant b \leqslant c$.
$$
\text { Let } b=a+s, c=a+s+t(s, t \geqslant 0) \text {. }
$$

Then by the condition,
$$
a^{2}+(a+s)^{2}+(a+s+t)^{2}=\lambda \text {. }
$$

Rearranging the above equation into a quadratic equation in terms of $a$,
$$
3 a^{2}+2(2 s+t) a+2 s^{2}+2 s t+t^{2}-\lambda=0 \text {. }
$$

Since the equation has a solution, we have,
$$
\begin{array}{l}
\Delta=4(2 s+t)^{2}-12\left(2 s^{2}+2 s t+t^{2}-\lambda\right) \geqslant 0 \\
\Rightarrow s^{2}+s t+t^{2} \leqslant \frac{3}{2} \lambda .
\end{array}
$$

Since the above inequality is symmetric in $s$ and $t$, we can assume $0 \leqslant s \leqslant t$,
$$
f=\min \left\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\right\}=s^{2} \text {. }
$$

Also, $\frac{3}{2} \lambda \geqslant s^{2}+s t+t^{2} \geqslant 3 s^{2}$, so $s^{2} \leqslant \frac{\lambda}{2}$. This equality holds if and only if $s=t=\sqrt{\frac{\lambda}{2}}$, i.e., $a=-\sqrt{\frac{\lambda}{2}}, b=0, c=\sqrt{\frac{\lambda}{2}}$. Therefore, $f_{\text {max }}=\frac{\lambda}{2}$.