Subspace help please

I'm just learning about subspaces and am confused as how to apply the 3 rules to determine if a subset is a subspace or not.
I need help with 2 questions, after that i think i should be fine.

{(x,y,z): x + y + z>=0}

and

{(x,y,z):(21x + y + 14z=0, 2x + 4y + z = 0}

thanks

1. It's easy to find a counter-example for this first one regarding scalar multiplication. Consider the following vector:

u = (1, 2, 3)

Certainly 1 + 2 + 3 ≥ 0

But consider a scalar,c, such that c = -1. Then:

cu = (-1)(1, 2, 3) = (-1, -2, -3)

Since cu is not an element of the subset (ie. -1 - 2 - 3 is clearly less than zero), the subset is not closed under scalar multiplication and it is not a subspace.

***********

2. It should be clear that this is closed under scalar multiplication. Assume that there exists a vector u = (x, y, z) that satisfies the above equations. Then, subbing in cu for some nonzero scalar c:

21(cx) + (cy) + 14(cz) = 0
c[21(x) + (y) + 14(z) = 0]
21(x) + (y) + 14(z) = 0

And similarly for the second equation. Thus the set is closed under scalar multiplication. Now, for scalar addition, assume that there exists two vectors u = (x, y, z) and v = (a, b, c) that satisfy the above equations. We want to see if u + v also satisfies the above equations, so sub that in:

21(x + a) + (y + b) + 14(z + c) = 0
21x + 21a + y + b + 14z + 14c = 0
(21x + y + 14z) + (21a + b + 14c) = 0

But since u and v satisfy these equations:

0 + 0 = 0
0 = 0

Thus the subset is closed under vector addition and it is a subspace.

Done!