(i) To find the number of possible seating arrangements for the 12 passengers out of 14 seats, we use permutations:

\(^{14}P_{12} = \frac{14!}{(14-12)!} = \frac{14!}{2!} = 4.36 \times 10^{10}\).

(ii) For the specific seating arrangement:

1. The 3 business people can be arranged in the front row in \(3! = 6\) ways.
2. The 5 students each sit at a window seat, which can be arranged in \(5! = 120\) ways.
3. Mr and Mrs Lin can sit in the same row on the same side of the aisle in \(3P2 \times 2 = 12\) ways (choosing 2 out of 3 rows and 2 ways to arrange them).
4. Mr and Mrs Brown can sit in another row on the same side of the aisle in \(2 \times 2 = 4\) ways.
5. Total arrangements = \(6 \times 120 \times 12 \times 4 = 17280\).