Answer: 75

Solution:

Let $a = \log_{10} x,$ $b = \log_{10} y,$ and $c = \log_{10} z.$ Taking logs of $xyz = 10^{81}$ gives $a + b + c = 81.$ Since $\log_{10} yz = b + c,$ the second condition is $a(b + c) + bc = ab + ac + bc$ $= 468.$

Squaring the sum, $$\begin{aligned} a^2 + b^2 + c^2 &= (a + b + c)^2 \\ &\quad {}- 2(ab + ac + bc) \\ &= 81^2 - 2 \cdot 468 \\ &= 6561 - 936 \\ &= 5625, \end{aligned}$$ so the requested value is $\sqrt{5625} = 75.$