Correct Answer: 3

Solution : $(a + b + c)^{2} = a^{2}  + b^{2} + c^{2} + 2(ab + bc + ca)$
$⇒(a + b + c)^{2} -3(ab + bc + ca) = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) –3(ab + bc + ca)$
Putting the values of $a, b,$ and $c$, we get,
$⇒(101 + 102 + 103)^{2}-3(101 \times 102 + 102 \times 103 + 103\times 101) = a^{2} + b^{2} + c^{2} - (ab + bc + ca)$
$⇒a^{2} + b^{2} + c^{2} -ab - bc - ca = 306^2 -3(10302 + 10506 + 10403)$
$⇒a^{2} + b^{2} + c^{2} -ab - bc - ca = 93636 -3(31211)$
$⇒a^{2} + b^{2} + c^{2} -ab - bc - ca = 93636 -93633$
$\therefore a^{2} + b^{2} + c^{2} -ab - bc - ca= 3$
Hence, the correct answer is 3.