Description for Correct answer:
We know that, \( \Large a^{3}+b^{3}+c^{3}-3abc \)

=\( \Large \left(a+b+c\right) \left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \)

=>\( \Large \left(a+b+c\right)= \left(\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}-ab-bc-ca}\right) \)

Given that,

\( \Large \frac{\left(2.247\right)^{3}+ \left(1.730\right)^{3}+ \left(1.023\right)^{3}-3 \times 2.247 \times 1.730 \times 1.023}{ \left(2.247\right)^{2}+ \left(1.730\right)^{2}+ \left(1.023\right)^{2}- \left(2.247 \times 1.730\right)- \left(1.730 \times 1.023\right)- \left(2.247 \times 1.023\right) } \)

=\( \Large \left(2.247+1.730+1.023\right) \)

=5.000 = 5