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