If $$\overrightarrow r .\overrightarrow a = \overrightarrow r .\overrightarrow b = \overrightarrow r .\overrightarrow c = \frac{1}{2}$$      for some non-zero vector $$\overrightarrow r ,$$ then the area of the triangle whose vertices are $$A\left( {\overrightarrow a } \right),\,B\left( {\overrightarrow b } \right)$$   and $$C\left( {\overrightarrow c } \right)$$  is ($$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $$   are non-coplanar)

Solution :
Any vector $$\overrightarrow r $$ can be represented in terms of three non-coplanar vectors $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c $$ as
$$\overrightarrow r = x\left( {\overrightarrow a \times \overrightarrow b } \right) + y\left( {\overrightarrow b \times \overrightarrow c } \right) + z\left( {\overrightarrow c \times \overrightarrow a } \right)......\left( {\text{i}} \right)$$
Taking dot product with $$\overrightarrow a ,\,\overrightarrow b $$  and $$\overrightarrow c ,$$ respectively, we have
$$x = \frac{{\overrightarrow r .\overrightarrow c }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}},\,y = \frac{{\overrightarrow r .\overrightarrow a }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}{\text{ and}}\,z = \frac{{\overrightarrow r .\overrightarrow b }}{{\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]}}$$
From $$\left( {\text{i}} \right),$$ we have
$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]\overrightarrow r = \frac{1}{2}\left( {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right)$$
$$\therefore $$ Area of $$\Delta ABC = \frac{1}{2}\left| {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right| = \left| {\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]\overrightarrow r } \right|$$