11. (18 points) Let $a_{1} , a_{2} , a_{3}$ form an arithmetic sequence, $a_{1}+a_{2}+a_{3}=15$; $b_{1} , b_{2} , b_{3}$ form a geometric sequence, $b_{1} b_{2} b_{3}$ $=27$. If $a_{1}+b_{1} , a_{2}+b_{2} , a_{3}+b_{3}$ are positive integers and form a geometric sequence, find the maximum value of $a_{3}$.
11. Let $a_{1}=5-d, a_{2}=5, a_{3}=5+d, b_{1}=$ $\frac{3}{q}, b_{2}=3, b_{3}=3 q$. Then, from the conditions, $5-d+\frac{3}{q}, 5+$ $d+3 q$ are both positive integers, and
$$
\left(5-d+\frac{3}{q}\right)(5+d+3 q)=64 \text {. }
$$

Therefore, the possible values of $5-d+\frac{3}{q}$ and $5+d+3 q$ are essentially only four pairs:
$(1,64),(2,32),(4,16)$, and $(8,8)$.
We will discuss the maximum value of $d$ for each pair.
(1) $\left\{\begin{array}{l}5-d+\frac{3}{q}=1, \\ 5+d+3 q=64\end{array} \Rightarrow 3 q+\frac{3}{q}=55\right.$
$\Rightarrow q_{1,2}=\frac{55 \pm 7 \sqrt{61}}{6} \Rightarrow d_{1,2}=4+\frac{3}{q_{1,2}}$.
The larger $d$ is clearly
$4+\frac{3}{\frac{55-7 \sqrt{61}}{6}}=4+\frac{55+7 \sqrt{61}}{2}$.
(2) $\left\{\begin{array}{l}5-d+\frac{3}{q}=2, \\ 5+d+3 q=32\end{array} \Rightarrow q+\frac{1}{q}=8\right.$
$\Rightarrow q_{1,2}=\frac{8 \pm \sqrt{64-4}}{2}=4 \pm \sqrt{15}$
$\Rightarrow d_{1,2}=3+\frac{3}{q_{1,2}}$.
The larger $d$ is clearly
$$
3+\frac{3}{4-\sqrt{15}}=15+3 \sqrt{15} .
$$