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B
31
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C
64
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D
32
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Solution
The correct option is B 31 Given, a+b+c=10anda2+b2+c2=38. Using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca, ⇒102=38+2(ab+bc+ca) ⇒2(ab+bc+ca)=102−38 ⇒(ab+bc+ca)=622=31.