Question 254828
 a^2 – bc = 7-------- 1
 b^2 + ac = 7-------- 2
 c^2 + ab = 7-------- 3

equating 1 and 2 
  then 2 and 3 
  then 3 and 1

we get either a=b+c
        or    b=c=-a

but if b=c=-a the given equations will not satisfy

so a=b+c

now add the three equations 1,2 & 3

we get  {{{a^2–bc+c^2+ab+b^2+ac=21}}}
=>   {{{a^2+b^2+c^2=21+bc-a(b+c)}}}

but since b+c=a

we get {{{a^2+b^2+c^2=21+bc-a^2}}}

from eq. 1 bc-a^2=-7

substituting this in above eq 

we get {{{a^2+b^2+c^2=21-7}}}

so  {{{a^2+b^2+c^2=14}}}  Ans.

ans.  a^2+b^2+c^2=14