In  $\Delta ABC,$D and E are points on BC and AC respectively such that BD=2DC and AE=3EC. Let P be the point of intersection of AD and BE. The ratio BP: PE has the value 

Consider the following diagram:      

There’s no loss of generality in assuming A to be the origin $\overrightarrow{0}$ , and B and C be the points $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively. By the section formula,
 $D\equiv \frac{\overrightarrow{b}+2\overrightarrow{c}}{3},E=\frac{3\overrightarrow{c}}{4}$                                   
The equations of AD can be written (in parametric form) as:
           
Similarly, the equation of BE can be written as
          .
AD and BE intersect at P. Thus, the position vector of P must satisfy the equations of both AD and BE. This means that we must have
                              
                        
Since $\overrightarrow{b}$ and $\overrightarrow{c}$ are non—collinear, we must have
     =0    and  =0
This system upon solving yields
Thus, the position vector of P can be obtained by substituting the value of in the equations for AD (or BE).
                                                
We now know the position vectors of B, P and E. We simply need to find BP:PE. Suppose this is m: 1. Then,
                         

Thus,            BP : PE = 8 : 3
The correct option is (C).