16.
定座標\(O(0,0,0),A(5,0,0),B(0,5,0),C(0,0,5)\),地面為\(ax+by+cz=0\),不失一般性,令\(a,b,c\ge0\)
\(\displaystyle\frac{5a}{\sqrt{a^2+b^2+c^2}}=3,\frac{5b}{\sqrt{a^2+b^2+c^2}}=2\),化簡可得\(3b^2=c^2\)
所求\(\displaystyle=\frac{5c}{\sqrt{a^2+b^2+c^2}}=\sqrt{3}\cdot2=2\sqrt{3}\)