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Question

If a, b, c are real numbers such that a+b+c=0 and a2+b2+c2=1, then (3a+5b−8c)2+(−8a+3b+5c)2+(5a−8b+3c)2 is equal to.

A
49
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B
98
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C
147
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D
294
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Solution

The correct option is A 147
(3a+5b8c)2=9a2+25b2+64c2+30ab80bc48ac

(8a+3b+5c)2=64a2+9b2+25c248ab+30bc+80ac

(5a8b+3c)2=25a2+64b2+9c280ab48bc+30ac

adding all three

=9+25+64+30(ab+bc+ca)80(ab+bc+ca)48(ab+bc+ca)

=9898(ab+bc+ca)

=9898(((a+b+c)2a2b2c2)/2)

=98+49

=147

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