Let's define the three numbers as follows:

• Let the first number be $x$.
• The second number is three more than twice the first, so it can be written as $2x + 3$.
• The third number is six more than the first, so it can be written as $x + 6$.

The sum of these three numbers is given as 57. Therefore, we can set up the following equation:

$x + (2x + 3) + (x + 6) = 57$

Now, combine like terms:

$x + 2x + 3 + x + 6 = 57$

Simplify:

$4x + 9 = 57$

Next, subtract 9 from both sides:

$4x = 48$

Now, divide both sides by 4:

$x = 12$

So, the first number is $12$.

Now that we know $x = 12$, we can find the other two numbers:

• The second number is $2x + 3 = 2(12) + 3 = 24 + 3 = 27$.
• The third number is $x + 6 = 12 + 6 = 18$.

Thus, the three numbers are 12, 27, and 18.

Would you like more details or have any questions on this solution?

Here are five related questions for further exploration:

1. What would happen if the sum of the numbers were different, say 60?
2. How would the solution change if the second number was 4 more than twice the first?
3. How can you check the solution once you have the values for the three numbers?
4. Can you solve a similar system where the sum is 100 and the relationships remain the same?
5. What are some real-world applications where you might encounter such problems?

Tip: Always define your variables clearly at the start of the problem, and carefully translate the given relationships into algebraic expressions to avoid confusion!