If `vecaxxvecb=vecbxxvecc=veccxxveca` then `veca+vecb+vecc=`

Prove that: [vecaxxvecb ,vecbxxvecc ,veccxxveca] = [veca vecb vecc]^2

If [vecaxxvecb vecbxxvecc veccxxveca]=lamda [veca vecb vecc]^2 then lamda is equal to (A) 1 (B) 2 (C) 3 (D) 0

If [vecaxxvecb vecbxxvecc veccxxveca]=lamda [veca vecb vecc]^2 then lamda is equal to (A) 1 (B) 2 (C) 3 (D) 0

Three points whose position vectors are veca,vecb,vecc will be collinear if (A) lamdaveca+muvecb=(lamda+mu)vecc (B) vecaxxvecb+vecbxxvecc+veccxxveca=0 (C) [veca vecb vecc]=0 (D) none of these

Three points whose position vectors are veca,vecb,vecc will be collinear if (A) lamdaveca+muvecb=(lamda+mu)vecc (B) vecaxxvecb+vecbxxvecc+veccxxveca=0 (C) [veca vecb vecc]=0 (D) none of these

If [vecaxxvecb vecbxxvecc veccxxveca]=lambda[veca vecb vecc]^2 , then lambda is equal to :

Let veca, vecb and vec c be any three vectors; then prove that [[vecaxxvecb, vecbxxvecc, veccxxveca]] = [[veca, vecb, vecc]]^2

If vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca) and [veca vecb vecc]=(1)/(3) , then x+y+z is equal to

If vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca) and [veca vecb vecc]=1/8 , then x+y+z=