\(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}=>\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x}{30}=\frac{3y}{60}=\frac{2x+3y-z}{30+60-28}=\frac{124}{62}=2\)
=> x=2.15=30
y=2.20=40
z=2.28=56