Step 1: Recall what determines basic strength of nitrogen compounds.
Basic strength depends on the availability of the lone pair on nitrogen for donation to a proton. Greater electron density on N means stronger base. Electron-withdrawing groups reduce basicity.
Step 2: Compare basicity of $NH_3$ and $N_2H_4$.
In $N_2H_4$, the second $-NH_2$ group exerts a mild $-I$ inductive effect on the other nitrogen, slightly reducing its electron density. So $NH_3 > N_2H_4$ in basicity.
Step 3: Compare basicity of $NH_2OH$ with $NH_3$ and $N_2H_4$.
In $NH_2OH$, highly electronegative oxygen is directly bonded to nitrogen. The strong $-I$ effect of oxygen significantly reduces electron density on N, making the lone pair much less available. Hence $NH_2OH$ is the weakest base.
Step 4: Establish the complete order from strongest to weakest.
$NH_3 > N_2H_4 > NH_2OH$. This matches option (3).
Step 5: Verify with $pK_b$ values.
$pK_b$: $NH_3 \approx 4.74$, $N_2H_4 \approx 5.77$, $NH_2OH \approx 8.04$. Lower $pK_b$ means stronger base. Order confirmed.
Step 6: State the final answer.
\[ \boxed{NH_3 > N_2H_4 > NH_2OH} \]