Your problem in understanding coplanarity is of distinguishing "points" from "vectors," and of distinguishing 2-dimensional linear subspaces from planes in general. Every three points determine a plane, as you say, but in general this plane doesn't pass through the origin. In linear algebra we single out the planes that pass through the origin, since they're subspaces of $\mathbb{R}^3$. Then we have a different definition of the plane determined by just two vectors, which is their linear span $\{au+bv: a, b \in \mathbb{R}\}$. This necessarily includes the origin. You can also think of the span of $u$ and $v$ in terms of points, as the plane determined by $u, v$, and the origin.
As you've said, if $w$ is in the span of $u$ and $v$, i.e. is a linear combination of them, then $u,v,w$ are linearly dependent. What this means is precisely that $u,v,w$ are coplanar, in the sense that $w$ is in the plane determined by $u,v,$ and the origin.
To show if two matrices are independent, you do exactly what you always do: if your matrices are $A$ and $B$, you want to show that $\alpha A+\beta B=0$ for $\alpha,\beta\in\mathbb{R}$ (or $\mathbb{C}$, depending) if and only if $\alpha=\beta=0$.
Best Answer
You're looking for such $\alpha, \beta, \gamma \in \mathbb{R}$ that $\alpha \neq 0$ or $\beta \neq 0$ or $\gamma \neq 0$ and:
$\displaystyle \alpha\begin{bmatrix}1 & 0 \\ 1 & 2\end{bmatrix}+\beta\begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}+\gamma\begin{bmatrix}0 & 1 \\ 2 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
If it's possible to find such $\alpha,\beta,\gamma$ matrices are linear dependent, if it's not possible are linear independent.
In this case first you look at first row and first column: it must be $\alpha=-\beta$, so:
$\displaystyle \alpha\begin{bmatrix}1 & 0 \\ 1 & 2\end{bmatrix}-\alpha\begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}+\gamma\begin{bmatrix}0 & 1 \\ 2 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
$\displaystyle \alpha\begin{bmatrix}0 & -2 \\ -1 & 1\end{bmatrix}+\gamma\begin{bmatrix}0 & 1 \\ 2 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$
Now look at second row second column, you must have $\alpha=-\gamma$, but when $\alpha=-\gamma$ you have:
$\displaystyle \alpha\begin{bmatrix}0 & -3 \\ -3 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$