Finding an unknown length in a triangle. I have incurred a question and I am having a hard time with it. Please refer the image below 
Here it is given that $AP:PC = 3:4$, $QM:MP = 3:2$ and $QB=12cm$. We have to find the length of $AB$. I have tried as many theorems I knew but was not able to conclude to the answer. Please help me to find the answer. Any help is appreciated. $$Thank you$$
 A: Menelaus' theorem for triangle $APQ$ and transversal line through $B,M,C$
$$\frac{BA}{BQ} \cdot \frac{MQ}{MP} \cdot \frac{CP}{CA}=1 \Rightarrow \frac{BA}{12} \cdot \frac{3}{2} \cdot \frac{CP}{CA}=1 \Rightarrow AB=8\cdot\frac{AC}{PC}=8\cdot\left(\frac{AP+PC}{PC}\right)$$
$$=8\cdot\left(\frac{AP}{PC}+1\right)=8\cdot\left(\frac{3}{4}+1\right)=14$$
A: In following link, you may see the triangle drawn.
http://i.imgur.com/AUskpyR.png

Solution:
$\frac{AP}{PC} = \frac{3}{4}, \frac{QM}{MP} = \frac{3}{2} $ are given in the question. Lets draw a line from A to T through M.
$\frac{AP}{AC} \times \frac{CT}{TQ} \times \frac{QM}{MP} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{3}{7} \times \frac{CT}{TQ} \times \frac{3}{2} = 1,
\frac{CT}{TQ} = \frac{14}{9}\tag{1} $$
\frac{QT}{QC} \times \frac{CP}{PA} \times \frac{AM}{MT} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{9}{23} \times \frac{4}{3} \times \frac{AM}{MT} = 1,
\frac{AM}{MT} = \frac{23}{12}\tag{2} $$
\frac{CT}{CQ} \times \frac{QB}{BA} \times \frac{AM}{MT} = 1$ by the menelaus theorem. When we plug the values into this:$
\frac{14}{23} \times \frac{12cm}{BA} \times \frac{23}{12} = 1,
\frac{12cm}{BA} = \frac{12}{14}\tag{3}$$Thus,  BA = 14 cm.$
