Given the subspaces $S=\{(1,2,1),(0,2,0)\}$ and $P=\{(x,y,z): x+y=y-kz=0\}$ 
Given the subspaces $S=\{(1,2,1),(0,2,0)\}$ and $P=\{(x,y,z): x+y=y-kz=0\}$, try to find $k$ such that: $S+P=\mathbb R^3$ and $S∩P={(0,0,0)}$.

Sorry, I forgot to put what I did try first. Here I go:
First, I became aware that if $S+P$ is equal to $P$, then there the elements on the basis of S plus the elements on the basis of $P$ correspond to the basis of $\mathbb R^3$.
But we know that $\mathbb R^3$ has a canonical basis, that is: {$(1,0,0), (0,1,0), (0,0,1)$}, and so now we can arrange $P$ so it can create this set.
This is my main idea, but I don't know how to continue.

You have a basis for $S$ (presuming $S$ is the span of the two vector set, and thus a subspace). Note that the vectors are linearly independent, as they are not scalar multiples of each other.
Now you need a basis for $P$. The equations $x + y = 0$ and $y - kz = 0$ simplify to $y = kz$ and $x = -y = -kz$. So, for $(x, y, z) \in P$,
$$(x, y, z) = (-kz, kz, z) = z(-k, k, 1).$$
Thus, every element of $P$ is a scalar multiple of $(-k, k, 1)$. You can also check that $(-k, k, 1)$ (and hence all of its multiples) belongs to $P$, so $\{(-k, k, 1)\}$ is a basis.
Now you can apply the result, and I'll leave the rest to you: for which values of $k$ is $\{(1, 2, 1), (0, 2, 0), (-k, k, 1)\}$ a basis for $\Bbb{R}^3$?
