Correct Answer: 2

Solution : $a+b+c=5$
Squaring both sides,
⇒ $(a+b+c)^2 = 25$
⇒ $a^2 + b^2 + c^2 +2(ab+bc+ca) = 25 $
⇒ $a^2 + b^2 + c^2 - ab - bc - ca + 3(ab+bc+ca) = 25$ ........(1)
Now we have to find the value of $(ab + bc + ca)$
$a+b+c=5$
Cubing both sides,
⇒ $a^3 + b^3 + c^3 + 3[(a+b+c)(ab+ac+bc)−abc] = 125$
⇒ $85 + 3[5 (ab+bc+ca) - 25] = 125$
⇒ $15(ab+bc+ca) - 75 = 40$
⇒ $3(ab+bc+ca) = \frac{115}{5}$
⇒ $3(ab+bc+ca) = 23$
Putting in (1),
⇒ $a^2 +b^2 + c^2 -ab - bc-ca = 25 -23$
⇒ $a^2 + b^2 +c^2 -ab - bc - ca = 2$
Hence, the correct answer is 2.