Cộng 3 vế lại ta được:
a+b+b+c+c+a=6+15+11
=> 2a+2b+2c=32
=> a+b+c=16
=> \(\hept{\begin{cases}a=\left(a+b+c\right)-\left(b+c\right)=16-15=1\\b=\left(a+b+c\right)-\left(c+a\right)=16-11=5\\c=\left(a+b+c\right)-\left(a+b\right)=16-6=10\end{cases}}\)
Vậy a=1,b=5,c=10