If a, b, c are real numbers such that a+b+c=0 | vedclass

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(C) Given $a+b+c=0$ and $a^2+b^2+c^2=1$.Let $S = (3a+5b-8c)^2+(-8a+3b+5c)^2+(5a-8b+3c)^2$.Expanding each term:$(3a+5b-8c)^2 = 9a^2+25b^2+64c^2+30ab-48

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$147$

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(C) Given $a+b+c=0$ and $a^2+b^2+c^2=1$.Let $S = (3a+5b-8c)^2+(-8a+3b+5c)^2+(5a-8b+3c)^2$.Expanding each term:$(3a+5b-8c)^2 = 9a^2+25b^2+64c^2+30ab-48ac-80bc$$(-8a+3b+5c)^2 = 64a^2+9b^2+25c^2-48ab-80ac+30bc$$(5a-8b+3c)^2 = 25a^2+64b^2+9c^2-80ab+30ac-48bc$Summing these expressions:$S = (9+64+25)a^2 + (25+9+64)b^2 + (64+25+9)c^2 + (30-48-80)ab + (-48-80+30)ac + (-80+30-48)bc$$S = 98a^2 + 98b^2 + 98c^2 - 98ab - 98ac - 98bc$$S = 98(a^2+b^2+c^2) - 98(ab+bc+ca)$Since $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = 0$,we have $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2) = -\frac{1}{2}(1) = -\frac{1}{2}$.Substituting these values:$S = 98(1) - 98(-\frac{1}{2}) = 98 + 49 = 147$.

If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$,then $(3a+5b-8c)^2+(-8a+3b+5c)^2+(5a-8b+3c)^2$ is equal to

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