Let \( \alpha, \beta \) be the roots of the equation \( x^2 - 3x + r = 0 \), and \( \frac{\alpha}{2}, 2\beta \) be the roots of the equation \( x^2 + 3x + r = 0 \). If the roots of the equation \( x^2 + 6x = m \) are \( 2\alpha + \beta + 2r \) and \( \alpha - 2\beta - \frac{r}{2} \), then \( m \) is equal to:
