14. (25 points) Given real numbers $a, b, c$ satisfy
$$
a^{2}+b^{2}+c^{2}=1 \text {. }
$$

Find the maximum and minimum values of $M=a^{2} b c+a b^{2} c+a b c^{2}$.
14. By the mean inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left|a^{2} b c+a b^{2} c+a b c^{2}\right| \\
\leqslant|a b c|(|a|+|b|+|c|) \\
\leqslant\left(\frac{|a|+|b|+|c|}{3}\right)^{3}(|a|+|b|+|c|) \\
\leqslant \frac{1}{27}\left(\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}\right)^{4}=\frac{1}{3},
\end{array}
$$

When $a=b=c=\frac{1}{\sqrt{3}}$, the equality holds in all the above inequalities.
Thus, the maximum value of $M$ is $\frac{1}{3}$.
To find the minimum value of $M$, we only need to consider the case where $a, b > 0$ and $c < 0$.
For a fixed $c = -t$, let
$$
a^{2}+b^{2}+t^{2}=1, \quad a+b-t>0,
$$

At this point, $t^{2} < \frac{2}{3}$, and $x = \frac{1}{3}$ is an extraneous root, so $x_{0} = \frac{1}{2} - \frac{\sqrt{2}}{4}$ is the maximum point of $f(x)$.
Therefore, $-f\left(x_{0}\right) = -\frac{1+\sqrt{2}}{16}$ is the minimum value of $M$.