We can think of the three particular people who must sit next to each other as forming a single "block" or group. This reduces the number of units to arrange. Here's how we can break down the problem:

1. Treat the 3 particular people as 1 block: So instead of arranging 6 people in 11 seats, we are arranging 4 units — this block (containing the 3 people) and the 3 other individuals.

2. Arrange the 4 units (block + 3 individuals): The 4 units can be arranged in the 11 available seats. Out of the 11 seats, we need to choose 4 positions where the units will be placed. The number of ways to do this is given by a combination formula: $\binom{11}{4} = \frac{11!}{4!(11-4)!} = 330$ So, there are 330 ways to select 4 seats.

3. Arrange the 3 people within the block: The 3 particular people within the block can be arranged among themselves in $3!$ (3 factorial) ways. That is: $3! = 6$

4. Arrange the remaining 3 individuals: The remaining 3 individuals can be arranged in the remaining 3 seats. This can be done in $3!$ ways: $3! = 6$

5. Final Calculation: To find the total number of ways, multiply all of these values together: $\text{Total ways} = 330 \times 6 \times 6 = 11,880$

Thus, the total number of ways to seat the 6 people such that the 3 particular people sit next to each other is 11,880.

Would you like a deeper explanation, or do you have any specific questions?

Here are 5 related questions to consider:

1. How would the solution change if the row had more or fewer seats?
2. What if there were more than 3 particular people who must sit together?
3. How many ways could 6 people be seated without any restrictions on seating?
4. How does the answer change if two specific people must not sit next to each other?
5. Can this method be generalized for different numbers of seats and people?

Tip: When particular people must sit together, treat them as a block first to simplify the arrangement process.