I think you're overthinking this. These three vectors lie in a common plane. A plane requires only two basis vectors to be completely described. Therefore, you can pick any two vectors that are linear combinations of the given set of 3, and you're done.
In other words, choose two basis vectors $b_1, b_2$ and express them as linear combinations of $v_1, v_2, v_3$. You have freedom to choose what $b_1, b_2$ are, provided that they are linearly independent of one another.
I personally feel that automatically putting things into "matrix form" and "row reducing" is too often a substitute for understanding what you are doing! A set of vectors, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$, is defined to be "linearly independent" if and only if the only solution to $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ is $a_1= a_2= \cdot\cdot\cdot= a_n= 0$.
The three vectors here are $(2, 2, -1)$, $(-8, -2, 5)$, $(-3, 0, 2)$ so the equation is $a(2, 2, -1)+ b(-8, -2, 5)+ c(-3, 0, 2)= (2a- 8b- 3c, 2a- 2b, -a+ 5b+ 2c)= (0, 0, 0)$ so we must have 2a- 8b- 3c= 0, 2a- 2b= 0, -a+ 5b+ 2c= 0. From 2a- 2b= 0, a= b. Putting that into the first equation, 2a- 8a- 3c= -6a- 3c= 0 so c= -2a. Putting b= a and c= -2a into the third equation -a+ 5a- 4a= 0a= 0. Then a(2, 2, -1)+ a(-8, -2, 5)- 2a(-3, 0, 2)= 0 so these vectors are linearly dependent. We can write a(2, 2, -1)= -a(-8, -2, 5)+ 2a(-3, 0, 2) and then divide by a: (2, 2, -1)= (-8, -2, 5)+ 2(-3, 0, 2). Since that vector can be written as a linear combination of the other two vectors, and those vectors [b]are[/b] independent (one is not a multiple of the other), the set {(-8, -2, 5), (-3, 0, 2)} is the largest subset of linearly independent vectors.
Best Answer
Set up the matrix $A$ whose rows are your vectors.
Use elementary row operations to bring it to reduced row-echelon form $B$.
Prove that the non-zero rows in $B$ are linearly independent and span the same subspace as is spanned by your original vectors.