Suppose $a$, $b$, and $c$ are relatively prime integers such that \[\frac{a}{b+c} = 2\qquad\text{and}\qquad \frac{b}{a+c} = 3.\] What is $|c|$?
1. Given the equations:
   \[
   \frac{a}{b+c} = 2 \quad \text{and} \quad \frac{b}{a+c} = 3
   \]
   we can rewrite these equations as:
   \[
   a = 2(b+c) \quad \text{and} \quad b = 3(a+c)
   \]

2. Substitute \( a = 2(b+c) \) into \( b = 3(a+c) \):
   \[
   b = 3(2(b+c) + c)
   \]
   Simplify the equation:
   \[
   b = 3(2b + 2c + c) = 3(2b + 3c) = 6b + 9c
   \]
   Rearrange the equation to isolate \( b \):
   \[
   b - 6b = 9c \implies -5b = 9c \implies b = -\frac{9}{5}c
   \]

3. Since \( a \), \( b \), and \( c \) are relatively prime integers, \( b = -\frac{9}{5}c \) implies that \( c \) must be a multiple of 5. Let \( c = 5k \) for some integer \( k \). Then:
   \[
   b = -9k
   \]

4. Substitute \( b = -9k \) and \( c = 5k \) back into \( a = 2(b+c) \):
   \[
   a = 2(-9k + 5k) = 2(-4k) = -8k
   \]

5. Now we have \( a = -8k \), \( b = -9k \), and \( c = 5k \). Since \( a \), \( b \), and \( c \) are relatively prime, the greatest common divisor (gcd) of \(-8k\), \(-9k\), and \(5k\) must be 1. This implies that \( k = \pm 1 \).

6. Therefore, \( c = 5k \) implies \( |c| = 5 \).

The final answer is \(\boxed{5}\)