$\dfrac{b+c+1}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b-3}{c}=\dfrac{1}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=2$
$\Rightarrow 2a=b+c+1; 2b=a+c+2; 2c=a+b-3; a+b+c=\dfrac 12$
$2a=b+c+1\\3a=a+b+c=1\\3a=\dfrac 32\\a=\dfrac12$
$2b=a+c+2\\3b=a+b+c+2\\3b=\dfrac 52\\b=\dfrac 56$
$a+b+c=\dfrac 12\Rightarrow c=-\dfrac 56$