#### Answer

$F_{AD}=900N$
$F_{AB}=693N$
$F_{AC}=646N$

#### Work Step by Step

We can determine the required forces as follows:
$F_{DA}=\frac{1}{3}F_{AD}\hat i-\frac{2}{3}F_{AD}\hat j+\frac{2}{3}F_{AD}\hat k$
$\implies \Sigma F_z=0$
$\implies \frac{2}{3}F_{AD}-600=0$
This simplifies to:
$F_{AD}=900N$
Now, the sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies F_{AB} cos 30-\frac{2}{3}F_{AD}=0$
$\implies F_{AB}cos 30-\frac{2}{3}(900)=0$
This simplifies to:
$F_{AB}=693N$
Now, the sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies \frac{1}{3}F_{AD}+F_{AB} sin30-F_{AC}=0$
We plug in the known values to obtain:
$\frac{1}{3}(900)+693 sin30-F_{AC}=0$
This simplifies to:
$F_{AC}=646N$