Answer
(i) $\vec{F}_3$ is directed along $-x$ axis.
(ii) $\vec{F}_3$ is directed along $-x$ axis.
(iii) $\vec{F}_3$ is directed along $-x$ axis and has a magnitude of $8000N$.
Work Step by Step
(i) Since the spacecraft is stationary, which means its acceleration equals $0$, we have $$\vec{F}_1+\vec{F}_2+\vec{F}_3=0 (1)$$
Since both $\vec{F}_1$ and $\vec{F}_2$ are directed along the $+x$ axis, for equation (1) to be possible, $\vec{F}_3$ has to be directed along the $-x$ axis.
(ii) Similarly, moving at a constant velocity means zero acceleration, which again leads to $$\vec{F}_1+\vec{F}_2+\vec{F}_3=0 $$
So $\vec{F}_3$ has to be directed along the $-x$ axis, too.
(iii) We have $\vec{F}_1=+3000N$ and $\vec{F}_2=+5000N$. Therefore, $$\vec{F}_3=0-\vec{F}_1-\vec{F}_2=-8000N$$
So $\vec{F}_3$ is directed along $-x$ and has a magnitude of $8000N$.
