Positive number x, y and z satisfy x y z=10^ 81 and ( _ 10 x)( _ 10...

\begin{aligned} (a+b+c)^2 & =a^2+b^2+c^2+2(a b+a c+b c) \\ \left(\log _{10} x\right)\left(\log _{10} y z\right) & =\left(\log _{10} x\right)\left(\log _{10} y+\log _{10} z\right) \\ & =\log _{10} x \log _{10} y+\log _{10} x \log _{10} z \\ \text { and } \quad x y z & =10^{81} \text { implies } \log _{10} x+\log _{10} y+\log _{10} z=81 \end{aligned} and $$x y z=10^{81} \text { implies } \log _{10} x+\log _{10} y+\log _{10} z=81$$ Using the property mentioned first, we get \begin{aligned} & \left(\log _{10} x\right)^2+\left(\log _{10} y\right)^2+\left(\log _{10} z\right)^2 \\ & =\left(\log _{10} x+\log _{10} y+\log _{10} z\right)^2-2\left(\log _{10} x \cdot \log _{10} y+\log _{10} x \cdot \log _{10} z+\log _{10} y \cdot \log _{10} z\right) \\ & \sqrt{\left(\log _{10} x\right)^2+\left(\log _{10} y\right)^2+\left(\log _{10} z\right)^2}=\sqrt{(81)^2-2 \times 468} \\ & \sqrt{\left(\log _{10} x\right)^2+\left(\log _{10} y\right)^2+\left(\log _{10} z\right)^2}=75 \end{aligned}