The dimension of a subspace is defined as the number of linearly independent vectors that span the subspace. To determine the dimension of the subspace spanned by the given vectors (-3, 0, 1), (1, 2, 1), and (3, 0, -1), we need to check if any of these vectors can be written as a linear combination of the others.

1. Verify Linear Independence:
First, we need to check if the three vectors are linearly independent. We can do this by setting up a linear system of equations and solving for the coefficients that make the linear combination equal to the zero vector.

Let's assume that the vectors can be expressed as linear combinations of each other:
(-3, 0, 1) = a(1, 2, 1) + b(3, 0, -1)

Expanding the equation, we get:
-3 = a + 3b
0 = 2a
1 = a - b

Solving this system of equations, we find that a = 0 and b = -1/3. Hence, the vectors are linearly independent.

2. Determine Dimension:
Since the three vectors are linearly independent, they form a basis for the subspace they span. The dimension of the subspace is therefore equal to the number of vectors in the basis, which in this case is 3.

However, the question asks for the dimension of the subspace, not the dimension of the vector space. The dimension of the vector space R3 is 3, so the dimension of the subspace must be less than or equal to 3.

To find the dimension of the subspace, we need to check if any of the vectors can be expressed as a linear combination of the others.

- The vector (-3, 0, 1) cannot be written as a linear combination of the other two vectors.
- The vector (1, 2, 1) cannot be written as a linear combination of the other two vectors.
- The vector (3, 0, -1) cannot be written as a linear combination of the other two vectors.

Since none of the vectors can be expressed as a linear combination of the others, the dimension of the subspace spanned by these vectors is 3.

However, this contradicts the given correct answer of '2'. Hence, the answer provided is incorrect.

In conclusion, the dimension of the subspace of R3 spanned by (-3, 0, 1), (1, 2, 1), and (3, 0, -1) is 3, not 2.