Lying in the same plane

Yes, they all lie on the plane: $4x-3y+2z=3$ Therefore they are all coplanar .

How I worked out the plane:

For any three points we can find a plane so we find the plane through the first three points and check if the fourth point is on this plane.

We can work out two vectors on the plane by subtracting pairs of points:

$\begin{pmatrix} 0\\ -1\\ 0\\ \end{pmatrix}-\begin{pmatrix} 2\\ 1\\ -1\\ \end{pmatrix}=\begin{pmatrix} -2\\ -2\\ 1\\ \end{pmatrix} ,\begin{pmatrix} 0\\ -1\\ 0\\ \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix} =\begin{pmatrix} -1\\ -2\\ -1\\ \end{pmatrix}$

We then find the crossproduct to find the normal vector:

$\begin{pmatrix} -2\\ -2\\ 1\\ \end{pmatrix} \times \begin{pmatrix} -1\\ -2\\ -1\\ \end{pmatrix}=\begin{pmatrix} 4\\ -3\\ 2\\ \end{pmatrix}$

So the plane is therefore (using one of the points):

$4x-3y+2z=d \Rightarrow 4(0)-3(-1)+2(0)=3=d$

So the plane is therefore:

$4x-3y+2z=3$

We can check if the third point lies on the plane:

$4(3)-3(3)+2(0)=3$

So the fourth point is on the same point as the first three so they are coplanar.

Equation of plane passing through 3 points (0,-1,0) (2,1,-1) (1,1,1) Matrix [X-0. Y+1. Z-0]. First row [2-0. 1+1. -1+0] second row [1-0. 1+1. 1-0]third row Solve the matrix equation of plane is 4x-3y+2z=3 Now four points are coplanar if 4th point (3,3,0) satisfies the above plane equation LHS:- 4 3-3 3=12-9=3=RHS