Given:
\( f(3x + 2y,\ 2x - 5y) = 19x \)
\( f(x,\ 2x) = 27 \)
Let \( a = 3x + 2y \) and \( b = 2x - 5y \). This means \( f(a,\ b) = 19x \).
We are also given \( f(x,\ 2x) = 27 \).
To match \( f(a,\ b) \) with the form \( f(x,\ 2x) \), we set the arguments such that \( a = kx \) and \( b = 2kx \) for some constant \( k \).
Specifically, we consider the case where \( a = \frac{19}{9}x \) and \( b = \frac{38}{9}x \), which implies \( b = 2a \).
We verify this transformation by solving for \( y \) in terms of \( x \) using \( 3x + 2y = \frac{19}{9}x \):
\( 2y = \frac{19}{9}x - 3x = \frac{19 - 27}{9}x = -\frac{8}{9}x \)
\( y = -\frac{4}{9}x \)
Substituting this \( y \) into \( 2x - 5y \):
\( 2x - 5\left(-\frac{4}{9}x\right) = 2x + \frac{20}{9}x = \frac{18 + 20}{9}x = \frac{38}{9}x \)
This confirms that if the first argument is \( \frac{19}{9}x \), the second argument is \( \frac{38}{9}x \), and the function value is \( 19x \).
Let \( u = \frac{19}{9}x \). Then \( x = \frac{9}{19}u \). The relationship \( b = 2a \) implies \( f(a,\ 2a) = 19x \).
Substituting \( u \) for \( a \) and \( x \) in terms of \( u \):
\( f(u,\ 2u) = 19 \left( \frac{9}{19}u \right) = 9u \)
We are given \( f(x,\ 2x) = 27 \). From our derived general form, \( f(x,\ 2x) = 9x \).
Equating these two expressions for \( f(x,\ 2x) \):
\( 9x = 27 \Rightarrow x = 3 \)
Therefore, the correct option is (A): 3.