If `veca, vecb and vecc` are any three non-coplanar vectors, then prove that points `l_(1)veca+ m_(1)vecb+ n_(1)vecc, l_(2)veca+m_(2)vecb+n_(2)vecc, l_(3)veca+m_(3)vecb+ n_(3)vecc, l_(4)veca + m_(4)vecb+ n_(4)vecc` are coplanar if `|{:(l_(1),, l_(2),,l_(3),,l_(4)),(m_(1),,m_(2),,m_(3),,m_(4)), (n_1,,n_2,, n_3,,n_4),(1,,1,,1,,1):}|=0`

Explanation

We know that, four points having position vectors, a,b,c and d are coplanar, if there exists scalarrs x,y,z and t such that
`xa+yb+zc+td=0` where x+y+z+t=0
So, the given points will be coplanar, if thre exists scalars x,y,z and t such that
`x(l_(1)a+m_(1)b+n_(1)c)+y(l_(2)a+m_(2)b+n_(2)c)+z(l_(3)a+m_(3)b+n_(3)c)+t(l_(4)a+m_(4)b+n_(4)c)=0`
where, `x+y+z+t=0`
`implies (l_(1)x+l_(2)y+l_(3)z+l_(4)t)a+(m_(1)x+m_(2)y+m_(3)z+m_(4)t)b+(n_(1)x+n_(2)y+n_(3)z+n_(4)t)c=0`
where, `x+y+z+t=0`
`l_(1)x+l_(2)y+l_(3)z+l_(4)t0` . . (i)
`m_(1)x+m_(2)y+m_(3)z+m_(4)t=0` . . (ii)
`n_(1)x+n_(2)y+n_(3)z+n_(4)t=0` . . (iii)
and x+y+z+t=0 . . (iv)
Eliminating x,y,z and t from above equation, we get
`|(l_(1),l_(2),l_(3),l_(4)),(m_(1),m_(2),m_(3),m_(4)),(n_(1),n_(2),n_(3),n_(4)),(1,1,1,1)|=0`