If xcosâ¡Î¸=ycosâ¡(θ+2Ï/3)=zcosâ¡(θ+4Ï/3) then xy+yz+zx=
cos2⡠θ
sin2⡠θ
1
0
K1x+1y+1z=cosâ¡Î¸+cosâ¡(θ+2Ï/3)+cosâ¡(θ+4Ï/3) =cosâ¡Î¸+2cosâ¡(θ+Ï)cosâ¡Ï/3=cosâ¡Î¸âcosâ¡Î¸=0â xy+yz+zx=0