Three non-collinear points are always define a plane. If fourth plane too is on this plane, four plane define this plane. So let us first define a plane using points A(3,-1,-1),B(-2,1,2)A(3,−1,−1),B(−2,1,2) and D(0,2,-1)D(0,2,−1), using vec(AB)=(B_x-A_x)hati+(B_y-A_y)hatj+(B_z-A_z)hatk−−→AB=(Bx−Ax)ˆi+(By−Ay)ˆj+(Bz−Az)ˆk
Therefore vec(AB)=(-2-3)hati+(1-(-1))hatj+(2-(-1))hatk=−−→AB=(−2−3)ˆi+(1−(−1))ˆj+(2−(−1))ˆk=
= -5hati+2hatj+3hatk−5ˆi+2ˆj+3ˆk and
vec(AD)=(0-3)hati+(2-(-1))hatj+(-1-(-1))hatk=−−→AD=(0−3)ˆi+(2−(−1))ˆj+(−1−(−1))ˆk=
= -3hati+3hatj+0hatk−3ˆi+3ˆj+0ˆk
If vec(AB)−−→AB and vec(AD)−−→AD are in the same plane, then we will have vec(AB)xxvec(AD)=0−−→AB×−−→AD=0, the cross product of the two vector as 00 and hence
|(hati,hatj,hatk),(-5,2,3),(-3,3,0)|=0
or (0-9)hati-(0-(-9))hatj+(-15-(-6))hatk=0
or -9hati-9hatj-9hatk=0
or hati+hatj+hatk=0
Hence equation of plane is x+y+z=k and putting values of points A,B and D, we get k=1
Hence equation of plane is x+y+z=1
and as C(8,-3,0) does not satisfy it,
A,B,C and D do not lie in the same plane.
