14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is
14.2.

From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$

which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqslant 2.
$$

When $a=2$, $b=c=-1$, which satisfies the problem's conditions.
Therefore, the maximum value of $a$ is 2.