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)# and #D(0,2,-1)#, using #vec(AB)=(B_x-A_x)hati+(B_y-A_y)hatj+(B_z-A_z)hatk#

Therefore #vec(AB)=(-2-3)hati+(1-(-1))hatj+(2-(-1))hatk=#

= #-5hati+2hatj+3hatk# and

#vec(AD)=(0-3)hati+(2-(-1))hatj+(-1-(-1))hatk=#

= #-3hati+3hatj+0hatk#

If #vec(AB)# and #vec(AD)# are in the same plane, then we will have #vec(AB)xxvec(AD)=0#, the cross product of the two vector as #0# 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.