If x is in the nullspace of A, then Ax = 0. Consider the augmented matrix
Add row 2 to row 1 to get this into reduced-row echelon form:
Then x = [x₁, x₂, x₃]ᵀ such that
x₁ - x₃ = 0
x₂ - 3 x₃ = 0
If you let x₃ = 1, you end up with x₁ = 1 and x₂ = 3, so x = [1, 3, 1]ᵀ.
More generally, if you let x₃ = y, then any vector of the form [y, 3y, y]ᵀ would belong to the nullspace.
Check that this is orthogonal to the rows of A :
[1, -1, 2] • [1, 3, 1]ᵀ = 1•1 + (-1)•3 + 2•1 = 1 - 3 + 2 = 0
[0, 1, -3] • [1, 3, 1]ᵀ = 0•1 + 1•3 + (-3)•1 = 0 + 3 - 3 = 0