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.
 A: The subspace $S$ generated by the two linearly independent vectors $(1,2,1)$ and $(0,2,0)$ is a plane and the subspace $Y$ of $\mathbb R^3$ defined by the two linearly independent equations $x+y=0$ and $y - kz = 0$, is of dimension $3-2=1$.
Since the sum of their dimensions equals the dimension of $\mathbb R^3$, your two conditions (on the sum and intersection) are equivalent. The easier to handle is the second one.
A vector of $S$, $$a(1,2,1)+b(0,2,0)=(a,2a+2b,a),$$ also belongs to $Y$ if and only if $3a+2b=0$ and $(2-k)a+2b = 0$, i.e. $b=-3a/2$ and $(-1-k)a=0$. Hence the condition for $S∩P$ to be $\{(0,0,0)\}$ is $-1-k\ne0$ i.e. $k\ne-1.$
A: 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 $\Bbb{R}^3$?
