Answer: D

Solution:

Subtract the second equation from the first: $(a-c)(b-1)=13.$ Put $u=a-c$ and $v=b-1.$ The four possibilities $(u,v)=(1,13),$ $(13,1),$ $(-1,-13),$ and $(-13,-1)$ give, after substitution into $bc+a=87,$ respectively $c=86/15,$ $c=74/3,$ $c=-8,$ and $c=100.$ Only the third is an integer solution of all three original equations: $(a,b,c)=(-9,-12,-8).$ (The last candidate is $(87,0,100),$ which fails $ca+b=60.$) Adding the original equations gives $ab+bc+ca$ ${}+(a+b+c)=247.$ Since $a+b+c=-29,$ we obtain $ab+bc+ca$ $=247-(-29)=276.$ Thus, D is the correct answer.