position vector of p = b +1/3 (b-c )  
                              = 4/3b - 1/3c
 position vector of q =  c + 1/3 ( c-a )
                              = 4/3 c - 1/3 a
                       of r =  4/3 a - 1/3 b
eq. of line AP,
        r = a + n1 ( 4/3b - 1/3c  -a )   = (1-n1 )a  + 4/3 n1 b - 1/3 n1 c   ........................1
line BQ , 
      r = b + n2 ( 4/3 c - 1/3 a  -b )   =   -1/3n2 a + (1- n2 )b  + 4/3 n2 c  ...............................2
 line CR ,
       r = c + n3 (4/3 a - 1/3 b  - c ) = 4/3 n3 a  -1/3 n3 b  + (1-n3 )c   .........................................3
 
position vector of X will be found by solving eq 1 &2 simultaneously , becoz it is the intersecting point of AP & BQ,
equating both eq.
(1-n1 )a  + 4/3 n1 b - 1/3 n1 c  =  -1/3n2 a + (1- n2 )b  + 4/3 n2 c
 comparing coefficients of a, b & c
 1-n1 = -1/3 n2     ,     4/3 n1  = (1- n2 )   ,   - 1/3 n1   =   4/3 n2
gives n1 = 12/13, so the position vector of X will be,              (putting n1 = 12/13 in eq 1 )
  x =  1/13 a + 16/13 b  -4/13 c .
 in the similar way we can solve eq 2 &3 & get Y   &  solve 1 & 3 to get Z
 Y =  1/13 b  + 16/13 c - 4/13 a
 Z = 1/13 c + 16/13 a  - 4/13 b
centroid of the triangle XYZ
           = (x + y +z )1/3
           = (1/13 + 16/13 - 4/13 ) ( a+b +c ) /3
           =  ( a+b +c ) /3
           , which is also the centroid of triangle ABC