21. (12 points) Given two moving points $A$ and $B$, and a fixed point $M\left(x_{0}, y_{0}\right)$, all on the parabola $y^{2}=2 p x (p>0)$ ($A$ and $B$ do not coincide with $M$). Let $F$ be the focus of the parabola, and $Q$ be a point on the axis of symmetry. It is given that $\left(\overrightarrow{Q A}+\frac{1}{2} \overrightarrow{A B}\right) \cdot \overrightarrow{A B}=0$, and $|\overrightarrow{F A}|, |\overrightarrow{F M}|, |\overrightarrow{F B}|$ form an arithmetic sequence.
(1) Find the coordinates of $\overrightarrow{O Q}$;
(2) If $|\overrightarrow{O Q}|=3, |\overrightarrow{F M}|=\frac{5}{2}$, and the projections of points $A$ and $B$ on the directrix of the parabola are $A_{1}$ and $B_{1}$ respectively, find the range of the area of quadrilateral $A B B_{1} A_{1}$.
21. (1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), Q(a, 0)$. Then, from $|\overrightarrow{F A}|, |\overrightarrow{F M}|, |\overrightarrow{F B}|$ forming an arithmetic sequence, we get $x_{0}=\frac{x_{1}+x_{2}}{2}$.
And $y_{1}^{2}=2 p x_{1}$,
$y_{2}^{2}=2 p x_{2}$.
(1) - (2) gives $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2 p}{y_{1}+y_{2}}$.

From $\left(\overrightarrow{Q A}+\frac{1}{2} \overrightarrow{A B}\right) \cdot \overrightarrow{A B}=0$, we get $a=x_{0}+p$. Hence, $\overrightarrow{O Q}=\left(x_{0}+p, 0\right)$.
(2) From $|\overrightarrow{O Q}|=3,|\overrightarrow{F M}|=\frac{5}{2}$, we get
$$
p=1, x_{0}=2 \text {. }
$$

Then $S_{\text {quadrilateral } A B B_{1} A_{1}}$
$$
\begin{array}{l}
=\frac{\left[\left(x_{1}+\frac{1}{2}\right)+\left(x_{2}+\frac{1}{2}\right)\right]\left|y_{1}-y_{2}\right|}{2} \\
=\frac{5}{2}\left|y_{1}-y_{2}\right| .
\end{array}
$$

Since $y_{1}= \pm \sqrt{2 x_{1}}, y_{2}= \pm \sqrt{2 x_{2}}$, we have
$$
y_{1} y_{2}= \pm 2 \sqrt{x_{1} x_{2}} \text {. }
$$

And $0 \leqslant 2 \sqrt{x_{1} x_{2}} \leqslant x_{1}+x_{2}=4$, thus,
$$
-4 \leqslant y_{1} y_{2} \leqslant 4 \text {. }
$$

Therefore, $0 \leqslant\left(y_{1}-y_{2}\right)^{2}=2\left(x_{1}+x_{2}\right)-2 y_{1} y_{2}$
$$
=8-2 y_{1} y_{2} \leqslant 16 \text {. }
$$

Then $S_{\text {quadrilateral } A B B_{1} A_{1}}=\frac{5}{2}\left|y_{1}-y_{2}\right| \in(0,10]$.
Thus, the range of the area of quadrilateral $A B B_{1} A_{1}$ is $(0,10]$.