Discussion

Explanation:

Given, $xyz$ = ${\left(\mathrm{xy}.\mathrm{z}\right)}^a$ = ${\left(x.y\mathrm{z}\right)}^b$ = ${\left(0.\mathrm{xyz}\right)}^c$

$xyz$ = ${\left(\mathrm{xy}.\mathrm{z}\right)}^a$ 
⇒ $xy.z$ = ${\left(\mathrm{xyz}\right)}^{1/a}$
⇒ $xyz\times 0.1$ = ${\left(\mathrm{xyz}\right)}^{1/a}$    ...(1)
⇒ 0.1 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}-1}$    ...(1)

$xyz$ = ${\left(x.y\mathrm{z}\right)}^b$ 
⇒ $x.yz$ = ${\left(\mathrm{xyz}\right)}^{1/b}$
⇒ $xyz\times 0.01$ = ${\left(\mathrm{xyz}\right)}^{1/b}$
⇒ 0.01 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{b}-1}$    ...(2)

$xyz$ = ${\left(0.\mathrm{xyz}\right)}^c$ 
⇒ $0.xyz$ = ${\left(\mathrm{xyz}\right)}^{1/c}$
⇒ $xyz\times 0.001$ = ${\left(\mathrm{xyz}\right)}^{1/c}$
⇒ 0.001 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$    ...(3)

We konw, (1) × (2) = (3)

⇒ 0.1 × 0.01 = 0.001

⇒  ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}-1}$ × ${\left(\mathrm{xyz}\right)}^{\frac{1}{b}-1}$ = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$

⇒  ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}+\frac{1}{b}-2}$ = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$

⇒ $\frac{1}{a}$ + $\frac{1}{b}$ - 2 = $\frac{1}{c}$ - 1

⇒ $\frac{1}{a}$ + $\frac{1}{b}$ = $\frac{1}{c}$ + 1

Hence, option (a)

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