Prove that `[veca+vecb, vecb+vecc, vecc+veca]=2[veca,vecb,vecc]`

If [vecaxx vecb vecb xx vecc vecc xx veca] = lamda [veca vecb vecc]^(2), then lamda is equal to :

[veca+2vecb-vecc,veca-vecb,veca-vecb-vecc] =

[veca+2vecb-vecc,veca-vecb,veca-vecb-vecc]=

If veca , vecb and vecc are three non-coplanar vectors and vecp , vecq and vecr are vectors defined by vecp = (vecb xx vecc)/([veca vecb vecc]) , vecq = (vecc xx veca)/([veca vecb vecc]) and vecr = (veca xx vec b)/([veca vecb vec c]) , then the value of (veca + vecb) * (vecb + vecc) * vecq + (vecc + vec a) * vecr =

For non zero vectors veca,vecb, vecc (veca xx vecb). Vecc= |veca| |vecb||vecc| holds iff:

If veca, vecb, vecc are three vectors such that veca + vecb + vecc= vec0 and |veca | =2, |vecb|=3, |vecc| = 5, then value of veca. vecb+vecb.vec c+vec c. veca is :

veca, vecb, vec c are three vectors such that veca + vecb + vec c = vec0, |veca| =1, |vecb| =2, |vec c|=3, then : veca.vecb + vecb .vecc+ vecc.veca is equal to :

If veca, vecb, vecc are three non-coplanar vectors, then [veca+ vecb + vecc veca - vecc veca-vecb] is equal to :

veca. (veca xx vecb)=