Let $$K=[\{\underbrace{(1,-3,2,-1)}_{x_1},\underbrace{(2,0,1,1)}_{x_2},\underbrace{(0,-2,1,-1)}_{x_3}\}],$$ $$L=[\{\underbrace{(1,-1,1,0)}_{y_1},\underbrace{(1,1,0,1)}_{y_2}\}],$$ $$M=[\{\underbrace{(2,3,1,-2)}_{z_1},\underbrace{(2,9,-2,1)}_{z_3}\}]$$ be subspaces of $\mathbb R^4.$ Find some base for $K\cap L\cap M.$
I got $K\leq L$ because: $$2y_1-y_2=x_1,\;y_1+y_2=x_2,\;y_1-y_2=x_3.$$
$L\& M:$ $$\;\alpha+\beta+2\gamma+2\delta=0$$ $$-\alpha+\beta+3\gamma+9\delta=0$$ $$\;\alpha+\;\;\;\;\;\;\;\;\gamma-2\delta=0$$ $$\;\;\;\;\;\;\;+\beta-2\gamma+\delta=0$$
$$\;\alpha+\beta+2\gamma+2\delta=0=I$$ $$\;\;\;\;2\beta+5\gamma+11\delta=0=II$$ $$\;\;\;\;\;\beta+\gamma+4\delta=0=III$$ $$\;\;\;\;\;\beta-2\gamma+\delta=0=IV$$
from $III\&IV:$ $$3\gamma+3\delta=0$$$$\implies \alpha+\beta=0$$$$\beta=3\gamma\implies \alpha=-3\gamma$$
so, for example: $-3y_1+3y_2+z_1=z_2\implies$ base for $L\cap M$ is $\{-3y_1+3y_2\}$ and, since $K\cap L=K$ can I write that, because $y_1-y_2\in K$, the base for $K\cap L\cap M$ is also $\{-3y_1+3y_2\}$?
We also have $L\subseteq K$, as $K$ is (at least) $2$ dimensional, since it contains two nonparallel vectors, and so is $L$, hence the containment $K\subseteq L$ can't be proper.
So $L=K$, and therefore indeed $K\cap L\cap M=L\cap M$.
If you happened to find a nonzero vector, such as $y_2-y_1=(0,2,-1,1)$, in the intersection $L\cap M$, then you can cut off to finish the proof simply by a dimension argument.