If the vertices of a triangle areA1,2,1,B4,3,1,C3,1,5 then  Area of Δ ABC42×cos2A=

The vertices of a triangle areA1,2,1,B4,3,1,C3,1,5The area of the triangle  ABCis12AB¯×AC¯HereAB¯=3i+j and AC¯=2i−j+4kHence the area of the triangle is 12ijk3102−14=12i4−j12+k−3−2=124i−12j−5kTherefore, the area of the triangle is  16+144+252=1852And cosA=AB¯⋅AC¯AB¯AC¯=6−19+14+1+16=51021=542            Consider  Area of Δ ABC42×cos2A=1852×42×542=18510