10. (20 points) Let $a_{1}, a_{2}, a_{3}$ be in arithmetic progression, $a_{1}+a_{2}+a_{3}=18$; $b_{1}, b_{2}, b_{3}$ be in geometric progression, $b_{1} b_{2} b_{3}=64$. If $a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}$ are positive integers and form a geometric progression, find the maximum value of $a_{3}$.
10. Let \(a_{1}=6-d, a_{2}=6, a_{3}=6+d\);
\[
b_{1}=\frac{4}{q}, b_{2}=4, b_{3}=4 q \text{. }
\]

Then, by the conditions, \(6-d+\frac{4}{q}\) and \(6+d+4 q\) are both positive integers; and
\[
\begin{array}{l}
\left(6-d+\frac{4}{q}\right)(6+d+4 q)=\left(a_{2}+b_{2}\right)^{2}=100 . \\
\text{ Hence }\left\{\begin{array}{l}
6-d+\frac{4}{q}=\frac{100}{t}, \\
6+d+4 q=t,
\end{array}\right.
\end{array}
\]
where \(t \in\{1,2,4,5,10,20,25,50,100\}\).
(1) + (2) gives
\[
4\left(q+\frac{1}{q}\right)=\frac{100}{t}+t-12 \text{. }
\]

From equation (2), \(a_{3}=6+d=t-4 q\).
Let \(f(t)=\frac{100}{t}+t-12\), and let the smaller root of \(4\left(q+\frac{1}{q}\right)=f(t)\) be \(q(t)\).

It is easy to see that \(f(t)\) is strictly increasing on \([10,+\infty]\), and as \(t\) decreases on \([10,+\infty]\), \(q(t)\) decreases.
From equation (4), when \(t=100\), \(a_{3}\) reaches its maximum value.
At this time, solving equation (3) gives \(q=\frac{89-9 \sqrt{97}}{8}\).
\[
\text{ Hence }\left(a_{3}\right)_{\max }=\frac{111+9 \sqrt{97}}{2} \text{. }
\]