The shortest distance between the two straight lines $\frac{x-4/3}{2}=\frac{y+6/5}{3}=\frac{z-3/2}{4}$ and $\frac{5y+6}{8}=\frac{2z-3}{9}=\frac{3x-4}{5}$ is

$\sqrt{29}$ 3 0 $6\sqrt{10}$

0

Since these two lines are intersecting, shortest distance between the lines will be 0. Hence (C) is the correct answer.