$\frac{\log P}{y-z} = \frac{\log Q}{z-x} = \frac{\log R}{x-y} = 10$ 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.