If 3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}, then the | vedclass

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(C) Given equation: $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}$Expanding the right side: $3a^{2}+3b^{2}+3c^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$Rearranging the te

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$a=b=c$

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(C) Given equation: $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2}$Expanding the right side: $3a^{2}+3b^{2}+3c^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$Rearranging the terms: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ac=0$Dividing by $2$: $a^{2}+b^{2}+c^{2}-ab-bc-ac=0$Multiplying by $2$: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ac=0$This can be rewritten as: $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$Since the sum of squares is zero,each term must be zero:$(a-b)^{2}=0 \Rightarrow a=b$$(b-c)^{2}=0 \Rightarrow b=c$$(c-a)^{2}=0 \Rightarrow c=a$Therefore,$a=b=c$.

If $3(a^{2}+b^{2}+c^{2})=(a+b+c)^{2},$ then the relation between $a, b$ and $c$ is

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