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Best answer
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$\frac{\log P}{y-z} = \frac{\log Q}{z-x} = \frac{\log R}{x-y} = 10$   for $x \neq y \neq z$

  • $\frac{\log P}{y-z} =10\implies\log P=10(y-z)\qquad \Rightarrow (1)$
  • $\frac{\log Q}{z-x} = 10\implies\log Q=10(z-x)\qquad \Rightarrow(2)$
  • $\frac{\log R}{x-y} = 10\implies\log R=10(x-y)\qquad \Rightarrow (3)$

Adding $(1),(2)$ and $(3)$ we get

$\log P+\log Q+\log R=10(y-z)+10(z-x)+10(x-y)$

$\implies \log P+\log Q+\log R=10\left(y-z+z-x+x-y\right)$

$\implies\log P+\log Q+\log R=10\times 0$

$\implies\log P+\log Q+\log R=0$

$\implies\log PQR=0$

$\implies PQR={10}^{0} = 1.$

Hence, (B) is the correct choice.

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$\log P= 10y – 10z \implies P = 10^{10y – 10z}$

$\log Q = 10z – 10x \implies Q = {10}^{10z – 10x }$

$\log R= 10x – 10y \implies R =  10^ {10x – 10y}$

$PQR = 10^{10x – 10y}* 10^{10z – 10x }* 10^ {10x – 10y}$

$\implies PQR = 10^{10y-10z+10z-10x+10x – 10y} = 10^0 = 1$
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