#### Answer

$s_{OA}=218mm$
$s_{OB}=327mm$

#### Work Step by Step

We can find the stretch in two of the springs as follows:
$F_{OC}=\frac{3}{7}F_{OC}\hat i+\frac{2}{7}F_{OC}\hat j+\frac{6}{7}F_{OC}\hat k$
The sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies \frac{3}{7}F_{OC}-F_{OB}=0$.eq(1)
The sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\frac{2}{7}F_{OC}-F_{OA}=0$.eq(2)
The sum of the forces in the z-direction is given as
$\Sigma F_z=0$
$\implies \frac{6}{7}F_{OC}-196.2=0$
This simplifies to:
$F_{OC}=229N$
We plug in the value of $F_{OC}$ in eq(1) to obtain:
$\frac{3}{7}(229)-F_{OB}=0$
This simplifies to:
$F_{OB}=98.1N$
We plug in the value of $F{OC}$ in eq(2) to obtain:
$F_{OA}=65.4N$
Now $s_{OB}=\frac{F_{OB}}{K}$
$\implies s_{OB}=\frac{98.1}{300}=327mm$
and $s_{OA}=\frac{F_{OA}}{K}$
$\implies s_{OA}=\frac{65.4}{300}$
$\implies s_{OA}=218mm$