Let \[ \frac{2x}{3} = \frac{4y}{5} = \frac{7z}{9} = k \] Then, \[ x = \frac{3k}{2}, \quad y = \frac{5k}{4}, \quad z = \frac{9k}{7} \] Now, \[ \frac{4x + 12y - 21z}{3y} = \frac{4 \times \frac{3k}{2} + 12 \times \frac{5k}{4} - 21 \times \frac{9k}{7}}{3 \times \frac{5k}{4}} = \frac{6k + 15k - 27k}{\frac{15k}{4}} = \frac{-6k}{\frac{15k}{4}} = -6k \times \frac{4}{15k} = -\frac{8}{5} \]