Let the three consecutive positive integers be \( a \), \( b \), and \( c \). Since they are consecutive integers, we can express them as:
\[ a = n, \quad b = n + 1, \quad c = n + 2 \]
for some positive integer \( n \).
According to the problem, we have the equation:
\[ b^2 = 4(a + c) \]
Substituting \( a \), \( b \), and \( c \) in terms of \( n \):
\[ (n + 1)^2 = 4(n + (n + 2)) \]
This simplifies to:
\[ (n + 1)^2 = 4(2n + 2) \]
Calculating the right side:
\[ (n + 1)^2 = 8n + 8 \]
Expanding the left side:
\[ n^2 + 2n + 1 = 8n + 8 \]
Rearranging this equation gives:
\[ n^2 + 2n + 1 - 8n - 8 = 0 \]
Simplifying further:
\[ n^2 - 6n - 7 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \( a = 1 \), \( b = -6 \), and \( c = -7 \):
\[ n = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \sqrt{36 + 28} = \sqrt{64} = 8 \]
Thus:
\[ n = \frac{6 \pm 8}{2} \]
Calculating the two possible values for \( n \):
\[ n = \frac{14}{2} = 7 \quad \text{or} \quad n = \frac{-2}{2} = -1 \]
Since \( n \) must be a positive integer, we take \( n = 7 \).
Now, substituting back to find \( c \):
\[ c = n + 2 = 7 + 2 = 9 \]
Thus, the value of \( c \) is:
\[ \boxed{9} \]