If `veca, vecb and vecc` are non-coplanar vectors, prove that the four points `2veca+3vecb-vecc, veca-2vecb+3vecc, 3veca+4vecb-2vecc and veca-6vecb+ 6 vecc` are coplanar.

Explanation

Let the given points be `A, B, C and D`. If they are coplanar, then the three coterminous vectors
`vec(AB), vec(AC) and vec(AD)` should be coplanar.
`" "vec(AB)=vec(OB)-vec(OA)=-veca-5vecb+4vecc`
`" "vec(AC) = vec(OC)-vec(OA)=veca+vecb-vecc`
and `" "vec(AD)=vec(OD)-vec(OA)=-veca-9vecb+7vecc`
Since the vectors `vec(AB), vec(AC), vec(AD)` are coplanar, we must have `|{:(-1,,-5,,4),(1,,1,,-1),(-1,,-9,,7):}|=0`, which is true.
Hence proved.