Given that:

$(a+b) : (b+c) : (c+a) = 2 : 3 : 4$

and

$a + b + c = 18$

we can assume the values of $a + b$, $b + c$, and $c + a$ in terms of a common ratio constant $k$ as follows:

1. $a + b = 2k$
2. $b + c = 3k$
3. $c + a = 4k$

Adding these three equations:

$(a + b) + (b + c) + (c + a) = 2k + 3k + 4k$

This simplifies to:

$2(a + b + c) = 9k$

Substituting $a + b + c = 18$:

$2 \times 18 = 9k$

$36 = 9k$

$k = 4$

Now we can substitute $k = 4$ back into the equations for $a + b$, $b + c$, and $c + a$:

1. $a + b = 2k = 2 \times 4 = 8$
2. $b + c = 3k = 3 \times 4 = 12$
3. $c + a = 4k = 4 \times 4 = 16$

Now, we have three equations:

1. $a + b = 8$
2. $b + c = 12$
3. $c + a = 16$

We can solve these equations by elimination. Subtract the first equation from the third:

$(c + a) - (a + b) = 16 - 8$

This simplifies to:

$c - b = 8 \quad \text{(Equation 4)}$

Next, subtract the second equation from the third:

$(c + a) - (b + c) = 16 - 12$

This simplifies to:

$a - b = 4 \quad \text{(Equation 5)}$

Now we have two new equations:

1. $c - b = 8$
2. $a - b = 4$

From $c - b = 8$, we get $c = b + 8$.

Using $a - b = 4$, we get $a = b + 4$.

Now substitute these values into $a + b + c = 18$:

$(b + 4) + b + (b + 8) = 18$

$3b + 12 = 18$

$3b = 6$

$b = 2$

Now substitute $b = 2$ back to find $a$ and $c$:

1. $a = b + 4 = 2 + 4 = 6$
2. $c = b + 8 = 2 + 8 = 10$

Thus, the value of $c$ is:

$\boxed{10}$

Would you like more details or have any questions?

Here are related questions you might find interesting:

1. How would the solution change if the ratio were different, such as $3 : 4 : 5$?
2. Can this method be applied to solve equations involving four variables with ratios?
3. What if $a + b + c$ was not known; how would you approach the problem?
4. How can we generalize this approach for ratios with more terms?
5. Can we solve this problem graphically by plotting the equations?

Tip: When dealing with ratios, assigning a common multiplier can simplify breaking down the relationships between variables.