Let |a¯|=|b¯|=|c¯|=1,(a¯,b¯)=(b¯,c¯)=(c¯,a¯)=π3 and xa¯+yb¯+zc¯=p(b¯×c¯)+q(c¯×a¯)+r(a¯×b¯) Also forms a right-handed system of vectors.  Then p+q+rx+y+z=

Given that |a|=|b|=|c|=1(a¯,b¯)=(b¯,c¯)=(c¯,a¯)=π3 And xa¯+yb¯+zc¯=p(b¯×c¯)+q(c¯×a¯)+r(a¯×b¯)⇒x+y2+z2=pabc=p2· since a¯,b¯,c¯2   =    a¯.a¯a¯.b¯a.¯c¯b¯.a¯b¯.b¯b¯.c¯c¯.a¯c¯.b¯c¯.a¯ =112121211212121 =12since a¯.a¯=1,    a¯.b¯ =1.1.cosπ3=12  , a¯.c¯=12               ∴2x+2y+z=2p→(1)similarly x+2y+z=2q→(2)and x+y+2z=2r→(3)(1)+(2)+(3)⇒4(x+y+z)=2(p+q+r)⇒p+q+rx+y+z=22