A) 75
B) 65
C) 60
D) 81
Correct Answer: D
Solution :
As we know, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] \[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)\] ?(i) and \[{{(a+b+c)}^{2}}={{11}^{2}}\] \[\therefore \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\,\,(ab+bc+ac)=121\] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\times 20=121\] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+40=121\] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=121-40=81\]
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