(12) Given a circle $O$ with radius $R$ and a line $l$ outside the circle, the projection of $O$ on $l$ is $H$, and $O H = d$. $O H$ intersects the circle $O$ at points $C$ and $D$, with $C H > D H$. Let $P$ and $Q$ be points on $l$ such that $P$ and $Q$ are on the same side of $H$, and $P H = a$, $Q H = b$, with $a > b$. In the circle $O$, there are 2008 chords $A_{i} B_{i} (i=1,2, \cdots, 2008)$ parallel to $l$, and these 2008 chords intersect $C D$ at points that divide $C D$ equally. Then,
$$
\frac{1}{2008} \sum_{i=1}^{2008}\left(P A_{i}^{2}+P B_{i}^{2}+Q A_{i}^{2}+Q B_{i}^{2}\right)
$$

is (expressed in terms of $a$, $b$, $d$, and $R$) $\qquad$
(12) As shown in the figure, let the midpoint of $A_{i} B_{i}$ be $M_{i}$, and the midpoint of $P Q$ be $N$. By the median formula, we have
$$
\begin{aligned}
& P A_{i}^{2}+P B_{i}^{2}+Q A_{i}^{2}+Q B_{i}^{2} \\
= & P Q^{2}+A_{i} B_{i}^{2}+4 M_{i} N^{2} \\
= & (a-b)^{2}+A_{i} B_{i}^{2}+4 M_{i} N^{2} .
\end{aligned}
$$

Since $M_{i}$ lies on $C H$, let $O M_{i}=x_{i}$, where we assume $x_{i}$ is positive when $M_{i}$ is on $O D$ and negative when $M_{i}$ is on $O C$. Then,
$$
\begin{aligned}
A_{i} B_{i}^{2}+4 M_{i} N^{2} & =A_{i} B_{i}^{2}+4\left(H N^{2}+M_{i} H^{2}\right) \\
& =4\left(R^{2}-x_{i}^{2}\right)+4\left[\left(\frac{a+b}{2}\right)^{2}+\left(d-x_{i}\right)^{2}\right] \\
& =(a+b)^{2}+4 R^{2}+4 d^{2}-8 d x_{i} .
\end{aligned}
$$

Since $\sum_{i=1}^{2008} x_{i}=0$, we have
$$
\begin{aligned}
& \frac{1}{2008} \sum_{i=1}^{2008}\left(P A_{i}^{2}+P B_{i}^{2}+Q A_{i}^{2}+Q B_{i}^{2}\right) \\
= & 2 a^{2}+2 b^{2}+4 d^{2}+4 R^{2} .
\end{aligned}
$$

Note: $P A_{i}^{2}+P B_{i}^{2}+Q A_{i}^{2}+Q B_{i}^{2}=P Q^{2}+$ $A_{i} B_{i}^{2}+4 M_{i} N^{2}$ is Euler's theorem.