#### Answer

$m=102Kg$

#### Work Step by Step

We can determine the required maximum mass as follows:
$F_{AD}=-\frac{2}{3}F_{AD}\hat i+\frac{2}{3}F_{AD}\hat j+\frac{1}{3}F_{AD}\hat k$
The sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies -\frac{2}{3}F_{AD}+F_{AB}=0$..eq(1)
The sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies \frac{2}{3}F_{AD}-F_{AC}=0$.eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{1}{3}F_{AD}-m(9.81)=0$.eq(3)
It is given that the maximum force is $3KN$, so let $F_{AD}=3KN$. Then we plug in this value in eq(1) and eq(2) to obtain:
$F_{AB}=F_{AC}=2KN$
Now we plug in the known values in eq(3) to obtain:
$\frac{1}{3}(3000)-m(9.81)=0$
This simplifies to:
$m=102Kg$