All the data given in the question points to the equation mentioned below:
Considering the three numbers to be a,b and c:
(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Accordingly, substitute for a, b and c and solve:
=> a2+b2+c2= 138, ab+bc+ac = 131. Thus, 2(ab+bc+ac)= 2×131= 262
(a+b+c)2= 138+262= 400.
Therefore, a+b+c= 20