For a set of 3 vectors to be a basis for R3, the vectors must be linearly independent.
That is, if a(3, -2, 0) + b(-7, 5, -3) + c(-2, 2, k) = (0, 0, 0), then it must be the case that a = b = c = 0.
We have the following system of equations:
3a - 7b - 2c = 0
-2a + 5b + 2c = 0
-3b + kc = 0
Add 2 times the first equation to 3 times the second and replace the second equation by the result:
3a - 7b - 2c = 0
b + 2c = 0
-3b + kc = 0
Add 3 times the second equation to the third and replace the third equation by the result:
3a - 7b - 2c = 0
b + 2c = 0
(6c + kc) = 0
So, c(6 + k) = 0
If k = -6, then c can be any real number.
If k ≠ -6, then 6 + k ≠ 0. So, c = 0, b = 0 and a = 0.
If k = -6, then the vectors do not form a linearly independent set. Otherwise, the vectors are not linearly independent and thus do not form a basis of R3.
