In $\Delta ABC$, $P$ is a point on $AB$ such that $PB:AP=3:4$ and $PQ$ is parallel to $AC$ (where $Q$ lies on $BC$). If $AR$ and $QS$ are perpendicular to $PC$ and $QS=9\text{cm}$, then what is the length of $AR$ (in $\text{cm}$)?

Given that $PQ\parallel AC$ in $\Delta ABC$, triangles $\Delta BPQ$ and $\Delta BAC$ are similar ($\Delta BPQ\sim \Delta BAC$).
1. From the given ratio, $PB:AP=3:4$, we can find the ratio of $BP$ to the entire side $BA$:

$$BA=BP+AP⟹\frac{BP}{BA}=\frac{3}{3+4}=\frac{3}{7}$$

Since $PQ\parallel AC$, by Thales' theorem (Basic Proportionality Theorem), we also have:

$$\frac{BQ}{BC}=\frac{3}{7}⟹\frac{BQ}{QC}=\frac{3}{4}$$

2. Now consider the areas of triangles sharing the same base or vertices:
* In $\Delta BPC$, $PQ$ is a line dividing the base $BC$ at $Q$. The ratio of the areas of $\Delta BPQ$ and $\Delta CPQ$ is equal to the ratio of their bases $BQ:QC$:

$$\frac{\text{Area}(\Delta BPQ)}{\text{Area}(\Delta CPQ)}=\frac{BQ}{QC}=\frac{3}{4}⟹\text{Area}(\Delta CPQ)=\frac{4}{3}\text{Area}(\Delta BPQ)$$

* Therefore, the total area of $\Delta BPC$ is:

$$\text{Area}(\Delta BPC)=\text{Area}(\Delta BPQ)+\text{Area}(\Delta CPQ)=\text{Area}(\Delta BPQ)+\frac{4}{3}\text{Area}(\Delta BPQ)=\frac{7}{3}\text{Area}(\Delta BPQ)$$

3. Similarly, for $\Delta ABC$, the line $PC$ divides the base $AB$ at $P$. The areas of $\Delta APC$ and $\Delta BPC$ are in the ratio of $AP:PB=4:3$:

$$\frac{\text{Area}(\Delta APC)}{\text{Area}(\Delta BPC)}=\frac{4}{3}⟹\text{Area}(\Delta APC)=\frac{4}{3}\times \left(\frac{7}{3}\text{Area}(\Delta BPQ)\right)=\frac{28}{9}\text{Area}(\Delta BPQ)$$

4. Now let's find the ratio of the areas of $\Delta APC$ and $\Delta CPQ$:

$$\frac{\text{Area}(\Delta APC)}{\text{Area}(\Delta CPQ)}=\frac{\frac{28}{9}\text{Area}(\Delta BPQ)}{\frac{4}{3}\text{Area}(\Delta BPQ)}=\frac{28}{9}\times \frac{3}{4}=\frac{7}{3}$$

5. Both $\Delta APC$ and $\Delta CPQ$ share the exact same base $PC$. The perpendiculars from opposite vertices $A$ and $Q$ to $PC$ are $AR$ and $QS$ respectively. Thus, the ratio of their areas is simply the ratio of these heights:

$$\frac{\text{Area}(\Delta APC)}{\text{Area}(\Delta CPQ)}=\frac{\frac{1}{2}\times PC\times AR}{\frac{1}{2}\times PC\times QS}=\frac{AR}{QS}$$

$$\frac{AR}{QS}=\frac{7}{3}$$

6. Given $QS=9\text{cm}$:

$$\frac{AR}{9}=\frac{7}{3}⟹AR=\frac{7\times 9}{3}=21\text{cm}$$