Question 14In triangle $ABC$, sides $a,b,c$ are opposite angles $A,B,C$, respectively. Given $\dfrac{b+c}{10}=\dfrac{c+a}{15}=\dfrac{a+b}{11}$, find $\sin A:\sin B:\sin C$.$2:3:4$$6:7:8$$8:3:7$$10:15:11$$21:25:26$