Check that the sets of points in $\mathbb R^3$ span the same affine subspace

Question

Check that the sets of points $\{(1, 1, 0) , (2, 1, 0) , (3, 1, 1)\}$ and $\{(2, 1, 0) , (3, 1 , 0) , (3, 1, 1)\}$ span in $\mathbb R^3$ the same affine subspace.

My try:
Consider the difference vectors of these sets.
For the first set: $(3,1,1)-(1,1,0)=(2,0,1), (2,1,0)-(1,1, 0)=(1,0,0)$
For the second set: $(3,1,1)-(2,1,0)=(1,0,1), (3,1,0)-(2,1,0)=(1,0,0) )$

It seems to me that since $(2,0,1)\neq (1,0,1)$ then these spaces span other parallelograms in an affine space and cannot span the same space, but I can't professionally justify it.

Answer 1

$Hint:$ The answer is positive if $(3,1,0)=c_1(1,1,0)+c_2(2,1,0)+c_3(3,1,1)$ holds for some real constants $c_i, (i=1,2,3).$

Answer 2

Hint: Refer to the book by David C Lay's linear algebra on section $1.4$ theorem $4$, the matrix $A =\begin{bmatrix} 1 & 2 & 3\\ 1 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$ can be row reduced in just one step so that the row echelon form has a pivot position in every row. The same is true for the other set. Also, you want to treat each point of your set as a vector and not point.