1. Let $a, b, c$ be real numbers, and satisfy
$$
a+b+c=15, a^{2}+b^{2}+c^{2}=100 \text {. }
$$

Then the product of the maximum and minimum values of $a$ is $\qquad$
$-1 . \frac{25}{3}$.
Since $100-a^{2}=b^{2}+c^{2}$
$$
\geqslant \frac{1}{2}(b+c)^{2}=\frac{1}{2}(15-a)^{2},
$$

Therefore, $3 a^{2}-30 a+25 \leqslant 0$.
Since the maximum and minimum values of $a$ are the two roots of the equation
$$
3 a^{2}-30 a+25=0
$$

and the maximum and minimum values of $a$ can be achieved, by Vieta's formulas, the product of the maximum and minimum values of $a$ is $\frac{25}{3}$.