Electrostatic Force!

$d^2$ = $(3 + 5)^2 + (0 - 4)^2 + (-2 - 1)^2$ = 89 $m^2$

c = 299792458 $ms^{-1}$

F = $\Large \frac{c^2 \times 10^{-7} \times 10 \mu \times 20 \mu \times 1000 m}{89}$ = (20.19674558959140764044943820224+) $mN$

Answer: $\boxed{20.197}$

Distance between the charges can be found by the Pythagorean Theorem:

$r = \sqrt{(3--5)^{2} + (0-4)^{2} + (-2-1)^{2}} = \sqrt{89}$

$Force = \frac{kQ_{1}Q_{2}}{r^{2}} = \frac{(8.9876{\times 10^{-6}})(20{\times 10^{-6}})(10{\times 10^{-6}})}{89} = 0.0202N$

We need to multiply by 1000 to convert from N to mN.

Hence, force = $\boxed{20.2}\ mN$

However, I'm reluctant to give the final answer to more than 2.s.f. because the data was only given to 2.s.f.

Vector components from point B to A corresponds to $R_{BA} = r_A - r_B = 3a_x - 2a_z - (-5a_x + 4a_y + a_z) = 8a_x - 4a_y - 3a_z$ . $F_{BA}=\frac{(10\mu C)(20\mu C)(R_{BA})}{4\pi\epsilon_0(|R_{BA}|)^3} = 17.127a_x - 8.563a_y - 6.423a_z \ \ mN$ $|F_{BA}|=\sqrt{(17.127)^2+(-8.563)^2+(-6.423)^2} \ \ mN = \boxed{20.197} \ \ mN$