If $ \overline a ,\overline b ,\overline c $ are position vectors of the vertices $ A,B,C $ of $ \Delta ABC. $ If $ \overline r $ is the position vector of a point $ P $ such that \[\left( {\left| {\overline b - \overline c } \right| + \left| {\overline c - \overline a } \right| + \left| {\overline a - \overline b } \right|} \right)\overline r = \left| {\overline b - \overline c } \right|\overline a + \left| {\overline c - \overline a } \right|\overline b + \left| {\overline a - \overline b } \right|\overline c \] then the point $ P $ is
A.Centroid of $ \Delta ABC $
B.Orthocentre of $ \Delta ABC $
C.Circumcentre of $ \Delta ABC $
D.Incentre of $ \Delta ABC $

Hint: For solving this particular question, we have to construct a triangle then we have to write the sides of the triangle in terms of position vector. After that we have to substitute the values in the given expression \[\left( {\left| {\overline b - \overline c } \right| + \left| {\overline c - \overline a } \right| + \left| {\overline a - \overline b } \right|} \right)\overline r = \left| {\overline b - \overline c } \right|\overline a + \left| {\overline c - \overline a } \right|\overline b + \left| {\overline a - \overline b } \right|\overline c \] .

Complete step-by-step answer:
Let us consider a $ \Delta ABC $ as shown in the given figure ,

Here vertex ‘A’ of the above $ \Delta ABC $ is equal to position vector $ \overline a $ ,
vertex ‘B’ of the above $ \Delta ABC $ is equal to position vector $ \overline b $ , and
vertex ‘C’ of the above $ \Delta ABC $ is equal to position vector $ \overline c $ .
Now, the side ‘AB’ of the $ \Delta ABC $ is equal to $ \left| {\overline a - \overline b } \right| $ ,
 the side ‘BC’ of the $ \Delta ABC $ is equal to $ \left| {\overline b - \overline c } \right| $ ,
the side ‘CA’ of the $ \Delta ABC $ is equal to $ \left| {\overline c - \overline a } \right| $ ,
It is given that $ \overline r $ is the position vector of a point $ P $ such that \[\left( {\left| {\overline b - \overline c } \right| + \left| {\overline c - \overline a } \right| + \left| {\overline a - \overline b } \right|} \right)\overline r = \left| {\overline b - \overline c } \right|\overline a + \left| {\overline c - \overline a } \right|\overline b + \left| {\overline a - \overline b } \right|\overline c \]
Now substitute the corresponding values, we will get ,
 \[
   \Rightarrow \left( {\left| {\overline b - \overline c } \right| + \left| {\overline c - \overline a } \right| + \left| {\overline a - \overline b } \right|} \right)\overline r = \left| {\overline b - \overline c } \right|\overline a + \left| {\overline c - \overline a } \right|\overline b + \left| {\overline a - \overline b } \right|\overline c \\
   \Rightarrow (BC + CA + AB)\overline r = BC\overline a + CA\overline b + AB\overline c \\
   \Rightarrow \overline r = \dfrac{{BC\overline a + CA\overline b + AB\overline c }}{{(AB + BC + CA)}} \\
 \]
where ‘AB’ , ‘BC’ , ‘CA’ represent the magnitude of the sides and \[\overline a ,{\text{ }}\overline b {\text{ and }}\overline c \] are the position vectors of ‘A’, ‘B’ and ‘C’ points respectively.
Here if we compare the equation of $ \overline r $ with the vector formula for incenter in a $ \Delta PQR\;{\text{where}}\;\overline p ,\;\overline q \;{\text{and}}\;\overline r $ are position vectors of vertices $ P,\;Q\;{\text{and}}\;R $ respectively is given as
Position vector of incenter $ = \dfrac{{QR.\overline p + RP.\overline q + PQ.\overline r }}{{PQ + QR + RP}},\;{\text{where}}\;PQ,\;QR\;{\text{and}}\;RP $ are magnitude of sides,
So we can see that the above equation is matching with the formula for incenter of a triangle,
Therefore, we can say that point ‘P’ is the incenter
Then,
We can say that ‘D’ is the correct option.
So, the correct answer is “Option C”.

Note: We must know that for a point say $ P $ to be Incentre we have the following expression ,
 \[\overline r = \dfrac{{AB\overline a + BC\overline b + CA\overline c }}{{(AB + BC + CA)}}\] . Where $ \overline r $ is the position vector of point $ P $ , where ‘AB’ , ‘BC’ , ‘CA’ represent the magnitude of the sides and \[\overline a ,{\text{ }}\overline b {\text{ and }}\overline c \] are the position vectors of ‘A’, ‘B’ and ‘C’ points respectively and the denominator \[AB + BC + CA\] is nothing but the perimeter of the given triangle.