.
To answer this question, you should organize the (4,3)-matrix comprised of given vectors as the rows,
and calculate its rank.
In fact, it can be done by using Gauss-Jordan elimination over the matrix rows, using elementary operations on rows of the matrix.
The full procedure is shown below step by step in all details.
The ANSWER is "YES" - these vectors span .
Your matrix
A1 A2 A3
1 1 -2 6
2 1 4 3
3 3 -3 1
4 2 2 3
Subtract the 1st row from the 2nd row and restore it
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 3 -3 1
4 2 2 3
Multiply the 1st row by 3
A1 A2 A3
1 3 -6 18
2 0 6 -3
3 3 -3 1
4 2 2 3
Subtract the 1st row from the 3rd row and restore it
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 3 -17
4 2 2 3
Multiply the 1st row by 2
A1 A2 A3
1 2 -4 12
2 0 6 -3
3 0 3 -17
4 2 2 3
Subtract the 1st row from the 4th row and restore it
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 3 -17
4 0 6 -9
Divide the 2nd row by 6
A1 A2 A3
1 1 -2 6
2 0 1 -1/2
3 0 3 -17
4 0 6 -9
Multiply the 2nd row by 3
A1 A2 A3
1 1 -2 6
2 0 3 -3/2
3 0 3 -17
4 0 6 -9
Subtract the 2nd row from the 3rd row and restore it
A1 A2 A3
1 1 -2 6
2 0 1 -1/2
3 0 0 -31/2
4 0 6 -9
Multiply the 2nd row by 6
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 -31/2
4 0 6 -9
Subtract the 2nd row from the 4th row and restore it
A1 A2 A3
1 1 -2 6
2 0 1 -1/2
3 0 0 -31/2
4 0 0 -6
Restore the 2nd row to the original view
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 -31/2
4 0 0 -6
Divide the 3rd row by -31/2
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 1
4 0 0 -6
Multiply the 3rd row by -6
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 -6
4 0 0 -6
Subtract the 3rd row from the 4th row and restore it
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 1
4 0 0 0
Restore the 3rd row to the original view
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 -31/2
4 0 0 0
Calculate the number of linearly independent rows
A1 A2 A3
1 1 -2 6
2 0 6 -3
3 0 0 -31/2
4 0 0 0
In good old times, students were supposed to do such calculations/transformations manually,
and I did it many times in my young students years at the university.
But now we live in the XXI century, and this tedious job can be done in seconds using computers and appropriate software.
This time, I used web-site www.reshish.com, which provides very robust, very convenient and very professional
online solver (on the free of charge basis).