Sorry the picture didn't come up. Here it is again.
The row reduction looks good to me. I'm not sure I would your matrix equation the way you did. I would write it this way:
Your column matrix
disappeared on the RHS for some reason. Other than that, as I say, it looks good to me!
[EDIT] See Deveno's post for a correction.
one must be extremely careful in accounting for each step of the row reduction.
i found that to eliminate λ from the row-reduced matrix, it was necessary to divide by λ. this suggests that λ = 0 is a special case.
and indeed, when λ = 0, we have 2(1,0,1,2) + 2(0,0,1,2) = (2,0,4,8), so that {(1,0,1,2), (2,0,4,8), (0,0,1,2)} is a linearly dependent set.
thus the statement: "regardless of the parameter λ, the only scalar values of c1, c2, c3 such that c1v1 + c2v2 + c3v3 = 0 are c1,c2,c3 all 0" is UNTRUE
i have exhibited a counter-example.