For each of the two given sets of three vectors in `RR^3`:
`S_1 = {(1 , 1 , 0) , (1 , 0 , 1) , (0 , 1 , 1)}`
`S_2 = {(-1 , 1 , 2) , (2 , -3 , 1) , (10 , -14 , 0)}`
Determine if the vectors in the set are linearly independent. If they are , the set is a basis for `RR^3`.

General Strategy for a Solution:
We can test the linear independence of a set of n vectors in `RR^n` using a system of linear equations or by calculating the determinant of the matrix formed by these vectors.

1. Form the Matrix: Construct a `bb{3 times 3}` matrix A , where the columns (or rows) are the given vectors.
2. Calculate the Determinant: Compute the determinant , det (A).
3.Draw the Conclusion:
If `bb {det(A) ne 0}` , the vectors are linearly independent , and the set is a basis.
If `bb {det(A) = 0}` , the vectors are linearly dependent , and the set is NOT a basis.

Specific theory used: A set B is a basis for a vector space V if B is linearly independent and B spans V.
Dimension Theorem: Since `dim(RR^3) = 3` , any set of 3 linearly independent vectors in `RR^3` is automatically a basis.
Determinant Test for Linear Independence: Let A be a square matrix whose columns are the vectors `v_1 , v_2 , ... , v_n`. The vectors are linearly independent if and only if `bb{det(A) ne 0}`.

Do the Exercise:
Case 1: Set `S_1 = {(1 , 1 , 0} , (1 , 0 , 1) , (0 , 1 , 1)}`.

1 . Form the Matrix: `A_1 = ([1 , 1 , 0] , [1 , 0 , 1] , [0 , 1 , 1])`.

2. Calculate the Determinant , `det (A_1)` , using cofactor expansion along the first row:

`det(A_1) = 1 * det([0 , 1] , [1 , 1]) -1 * ([1 , 1] , [0 , 1]) + 0 * ([1 , 0] , [0 , 1]) =`

`1 * (0 * 1 - 1 * 1) - 1 * (1 * 1 - 0 * 1) + 0 = -1 - 1 = -2`.

3. Conclusion for `S_1`: Since `det(A_1) = -2 ne 0` , the vectors in `S_1` are linearly independent. Therefor `bb{S_1}` is a basis for `RR^3` over `RR`. OK

Case 2: Set `S_2 = {(-1 , 1 , 2) , (2 , -3 , 1) , (10 , -14 , 0)}`.

1. Form the Matrix: `A_2 = ([-1 , 2 , 10] , [1 , -3 , -14] , [2 , 1 , 0])`.

2. Calculate the Determinant , `det(A_2)` , using cofactor expansion along the third row for simplicity:

`det (A_2) = 2 * det([2 , 10] , [-3 , -14]) - 1 * ([-1 , 10] , [1 , -14]) + 0 * ([-1 , 2] , [1 , -3]) =`

`2 * (2 * (-14) - (-3) * 10) -1 * ((-1) * (-14) - 1 * 10) + 0 = 4 - 4 = 0`.

3. Conclusion for `S_2`: Since `det (A_2) = 0` , the vectors in `S_2` are linearly dependent. Therefore , `bb{S_2}` is NOT a basis for `RR^2` over `RR`. END