In an ABCD.EFGH cuboid with AB = 4 cm, BC = 3 cm, and CG = 5 cm there is a parallelogram OBFPH with O is located at the center of ABCD and P is located at the center of EFGH. The distance between the lines HO and PB is ....

A. $\displaystyle 5\sqrt3$ cm

B. $\displaystyle 5\sqrt2$ cm

C. $\displaystyle \sqrt5$ cm

D. $\displaystyle \frac{5}{2}\sqrt2$ cm

E. $\displaystyle \frac{5}{3}\sqrt3$ cm

By making use of the parallelogram formula, I got $\displaystyle \frac{1}{5}\sqrt5$. Do you guys get the same answer as me or any of the options?