5.12 Let real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy the conditions
$$
\left\{\begin{array}{l}
x_{1}^{2}+\cdots+x_{6}^{2}=6 \\
x_{1}+\cdots+x_{6}=0 .
\end{array}\right.
$$

Find the maximum possible value of $x_{1} x_{2} \cdots x_{6}$.
Let the number of negative numbers among $x_{1}, \cdots, x_{6}$ be $a$. From (1) and (2), we know $a \notin\{0,6\}$. If $a$ is odd, then $x_{1} \cdots x_{6} \leqslant 0$; if $a$ is even, $a \in\{2,4\}$. We only need to consider the case $a=4$ (otherwise, replace $x_{i}$ with $-x_{i}$ for discussion). Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant 0,0>x_{3} \geqslant \cdots \geqslant x_{6}$. From (2), we know $x_{1}+x_{2}=$ $-\left(x_{3}+\cdots+x_{6}\right)$, let $k=x_{1}+x_{2}$. By the Cauchy-Schwarz inequality, we have
$$
\left(x_{1}+x_{2}\right)^{2} \leqslant 2\left(x_{1}^{2}+x_{2}^{2}\right), \frac{1}{2}\left(x_{3}+\cdots+x_{6}\right)^{2} \leqslant 2\left(x_{3}^{2}+\cdots+x_{6}^{2}\right) .
$$

Adding the two inequalities, we get $\frac{3}{2} k^{2} \leqslant 2\left(x_{1}^{2}+\cdots+x_{6}^{2}\right)=12$, hence $k \leqslant 2 \sqrt{2}$. Using this result and the AM-GM inequality, we have
$$
x_{1} x_{2} \cdots x_{6} \leqslant\left(\frac{x_{1}+x_{2}}{2}\right)^{2}\left(\frac{-x_{3}-\cdots-x_{6}}{4}\right)^{4}=\frac{k^{6}}{4 \cdot 4^{4}} \leqslant \frac{(2 \sqrt{2})^{6}}{4^{5}}=\frac{1}{2} .
$$

Equality holds when $x_{1}-x_{2}-\sqrt{2}, x_{3}-\cdots-x_{6}--\frac{\sqrt{2}}{2}$. Therefore, the maximum value of $x_{1} x_{2} \cdots x_{6}$ is $\frac{1}{2}$.