If $x^{3 a} = y^{2 b} = z^{4 c} = x y z,$ then find the value of $3 a b + 4 b c + 6 c a$ :

3 a b c

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8 a b c

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9 a b c

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12 a b c

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Solution

The correct option is B $12 a b c$
It is given that $x^{3 a} = y^{2 b} = z^{4 c} = x y z$

Writing $y$ and $z$ in terms of $x$ using above relation

$y = x^{\frac{3 a}{2 b}}$ , and $z = x^{\frac{3 a}{4 c}}$

Now, $x y z = x \times x^{\frac{3 a}{2 b}} \times x^{\frac{3 a}{4 c}} = x^{1 + \frac{3 a}{2 b} + \frac{3 a}{4 c}} = x^{\frac{4 b c + 6 a c + 3 a b}{4 b c}} . . . . . . . (1)$

From given relation $x y z = x^{3 a} . . . . . . . . (2)$

From (1) and (2)

$x^{\frac{4 b c + 6 a c + 3 a b}{4 b c}} = x^{\frac{3 a}{1}} \Rightarrow \frac{4 b c + 6 a c + 3 a b}{4 b c} = 3 a \Rightarrow 4 b c + 6 a c + 3 a b = 12 a b c$

Hence, $3 a b + 4 b c + 6 c a = 12 a b c$ .

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