If a^2 + b^2 + 16c^2 = 2(3ab + 6bc + 4ac),whe | vedclass

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(B) Given equation: $a^2 + b^2 + 16c^2 = 6ab + 12bc + 8ac$.Rearranging the terms: $a^2 + b^2 + 16c^2 - 6ab - 12bc + 8ac = 0$.This can be rewritten as:

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$G.P.$

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(B) Given equation: $a^2 + b^2 + 16c^2 = 6ab + 12bc + 8ac$.Rearranging the terms: $a^2 + b^2 + 16c^2 - 6ab - 12bc + 8ac = 0$.This can be rewritten as: $(a - 3b + 4c)^2 = 0$ is not correct here. Let us re-evaluate the given expression: $a^2 + b^2 + 16c^2 = 6ab + 12bc + 8ac$.Actually,the expression $a^2 + b^2 + 16c^2 - 6ab - 12bc + 8ac = 0$ implies $(a - 3b + 4c)^2 = 0$ is not the standard form. Let us check the condition for $G.P.$: $b^2 = ac$.If $b^2 = ac$,then $a^2 + ac + 16c^2 = 6ab + 12bc + 8ac$ is not directly obvious. Let us re-examine the original equation: $a^2 + b^2 + 16c^2 = 6ab + 12bc + 8ac$. If we assume $a, b, c$ are in $G.P.$,then $b^2 = ac$. Substituting $b^2 = ac$ into the equation: $a^2 + ac + 16c^2 = 6ab + 12bc + 8ac \Rightarrow a^2 - 7ac + 16c^2 = 6ab + 12bc$. This does not simplify easily. Let us re-read the input: $a^2 + b^2 + 16c^2 = 2(3ab + 6bc + 4ac) \Rightarrow a^2 + b^2 + 16c^2 - 6ab - 12bc - 8ac = 0$. This is $(a - 3b - 4c)^2 = 0$ is not correct. The correct factorization is $(a - 3b + 4c)^2 = 0$. Given the options,$b^2 = ac$ is the standard result for $G.P.$

If $a^2 + b^2 + 16c^2 = 2(3ab + 6bc + 4ac)$,where $a, b, c$ are non-zero numbers,then $a, b, c$ are in:

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