If the vectors `veca, vecb, and vecc` are coplanar show that `|(veca,vecb,vecc),(veca.veca, veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|=0`

Explanation

Given that `veca, vecb and vecc` are three coplanar vectors. Therefore, there exist scallars x,yand z , we all zone such that
`x veca + yvecb + zvecc = vec0`
Taking dot product of `veca and ` (i), we get
`xveca + yvecb +zvecc=vec0`
Again taking dot product of `vecb and` (i) we get
`xvecb.veca +yvecb.vecb+z vecb.vecc=0`
(i) Now Eqs. (i), (ii) and (iii) form a homogenecous system of equations , where, x , y and z are not all zero, therefore the systems must have a non-trival solution. and for this , the determinant of coefficient matix should be zero , i.e.,
`|{:(veca,vecb,vecc),(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc):}|=0`