11 (18 points) Let $a_{1}, a_{2}, a_{3}$ be in an arithmetic sequence, $a_{1}+a_{2}+a_{3}=15$; $b_{1}, b_{2}, b_{3}$ be in 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 we know that $5-d+\frac{3}{q}$ and $5+d+3 q$ are both positive integers and
$$
\left(5-d+\frac{3}{q}\right)(5+d+3 q)=64 \text {. }
$$

Therefore, the possible values for $5-d+\frac{3}{q}$ and $5+d+3 q$ are essentially four pairs: $(1,64)$, $(2,32)$, $(4,16)$, and $(8,8)$.
(1) For $(1,64)$, solving the system of equations $\left\{\begin{array}{l}5-d+\frac{3}{q}=1 \text {, } \\ 5+d+3 q=64,\end{array}\right.$ yields $3 q+\frac{3}{q}=$ 55, thus we have $q_{1,2}=\frac{55 \pm 7 \sqrt{61}}{6}$, and at this point $d_{1,2}=4+\frac{3}{q_{1,2}}$, the larger one is clearly
$$
4+\frac{3}{\frac{55-7 \sqrt{61}}{6}}=4+\frac{55+7 \sqrt{61}}{2} .
$$
(2) For $(2,32)$, solving the system of equations $\left\{\begin{array}{l}5-d+\frac{3}{q}=2 \text {, } \\ 5+d+3 q=32,\end{array}\right.$ yields $q+\frac{1}{q}=8$,

thus we have
$$
q_{1,2}=\frac{8 \pm \sqrt{64-4}}{2}=4 \pm \sqrt{15},
$$

at this point $d_{1,2}=3+\frac{3}{q_{1,2}}$, the larger one is clearly
$$
3+\frac{3}{4-\sqrt{15}}=15+3 \sqrt{15} \text {. }
$$
(3) For $(4,16)$, solving the system of equations $\left\{\begin{array}{l}5-d+\frac{3}{q}=4 \text {, } \\ 5+d+3 q=16,\end{array}\right.$ yields $3 q+\frac{3}{q}=$ 10, thus $q=3$ or $\frac{1}{3}$, at this point $d_{1,2}=1+\frac{3}{q_{1,2}}=2$ or 10, so the larger one is 10.
(4) For $(8,8)$, solving the system of equations $\left\{\begin{array}{l}5-d+\frac{3}{q}=8, \\ 5+d+3 q=8,\end{array}\right.$ yields $q+\frac{1}{q}=2$, thus $q=1$, and hence $d=0$.

Considering all four cases, the maximum value of $d$ is $4+\frac{55+7 \sqrt{61}}{2}$, so the largest $a_{3}$ is $9+\frac{55+7 \sqrt{61}}{2}$.