i. If `veca, vecb and vecc` are non-coplanar vectors, prove that vectors `3veca-7vecb-4vecc, 3veca-2vecb+vecc and veca+vecb+2vecc` are coplanar.

Explanation

Given `[veca vecb vecc] ne 0 as veca, vecb, vecc` are non- coplanar. Also there does not exist any linear relation between them because if any such relation extists, then they would be coplanar.
`Let " " A=x(vecbxxvecc)+y(vecc xxveca) + z (vecaxxvecb),`
`where " " x=veca.vecd,y=vecb.vecd,x =vecc.vecd`
we have to find value of modulus of `vecA i.e. |vecA|` which is independent of `vecd`.
Multiplying both sides scalarly by ` veca,vecb and vecc`. and knowing that scalar triple product is zero when two vectors are equal, we get
`vecA.veca=x[vecavecbvecc]+0`
putting for x, we get
`(veca.vecd)[vecavecb vecc]=vecA.veca`
similarly , we have
`(vecb.vecd)[vecavecbvecc]=vecA.vecb`
`(vecc.vecd)[veca vecbvecc]=vecA.vecc`
Adding the above relations , we get
`[(veca+vecb+vecc).vecd][vecavecbvecc]=vecA.(veca+vecb+vecc)`
`(veca+vecb+vecc).[vecd[veca vecbvecc]-A vecA=0`
since `veca vecb and vecc` are non-coplanar, `veca+vecb+vecc ne 0` because otherwise any one is expressible as a linear combination of other two, Hence,
`[veca vecbvecc] vecd=vecA`
`|vecA|=|[vecavecbvecc]|as vecd` is a unit vector.
It is independent of `vecd`.