11. (22 points) Let $a, b, c$ be real numbers not all zero. Find the range of $F=\frac{a b-b c+c^{2}}{a^{2}+2 b^{2}+3 c^{2}}$. Under what conditions on $a, b, c$ does $F$ achieve its maximum and minimum values?
11. This problem is to find a real number $F$ such that the equation
$$
F a^{2}+2 F b^{2}+(3 F-1) c^{2}-a b+b c=0
$$

has a real solution $(a, b, c) \neq(0,0,0)$.
When $F=0$, equation (1) clearly has a solution
$$
(a, b, c) \neq(0,0,0).
$$

When $F \neq 0$, equation (1) has a solution $(a, b, c) \neq(0,0,0)$ if and only if the inequality
$$
(c-a)^{2}-8 F\left[F a^{2}+(3 F-1) c^{2}\right] \geqslant 0
$$

has a real solution $(a, c) \neq(0,0)$.
Inequality (2) has a solution $(a, c) \neq(0,0)$ if and only if
$$
8 F^{2}-1 \leqslant 0
$$

or $\left(8 F^{2}-1\right)\left(24 F^{2}-8 F-1\right)-1 \leqslant 0$.
Notice that,
$$
\begin{array}{l}
\left(8 F^{2}-1\right)\left(24 F^{2}-8 F-1\right)-1 \\
=8 F(2 F-1)\left(12 F^{2}+2 F-1\right),
\end{array}
$$

and $\frac{-1-\sqrt{13}}{12}<-\frac{1}{\sqrt{8}}<0<\frac{-1+\sqrt{13}}{12}<\frac{1}{\sqrt{8}}<\frac{1}{2}$.
Inequality (3) holds if and only if $\frac{-1-\sqrt{13}}{12} \leqslant F \leqslant \frac{1}{2}$.
At this time, $c=\left(1-8 F^{2}\right) a, b=\frac{a-c}{4 F}=2 F a$.
In particular, when $b=a, c=-a$, $F$ takes the maximum value $\frac{1}{2}$; when $b=\frac{-1-\sqrt{13}}{6} a, c=\frac{2-\sqrt{13}}{9} a$, $F$ takes the minimum value $\frac{-1-\sqrt{13}}{12}$.