1) Let's denote points $A,B$ and $C$. Let's assume they are collinear, then $\overrightarrow{\pmb {AB}}$ and $\overrightarrow{\pmb {AC}}$ are collinear as well, so $\overrightarrow{\pmb {AB}} = k \overrightarrow{\pmb {AC}}$. On the other hand $\overrightarrow{\pmb {AB}} = \pmb b - \pmb a$ and $\overrightarrow{\pmb {AC}} = \pmb c - \pmb a$, so $\pmb b - \pmb a = k (\pmb c - \pmb a)$ which means $(k+1)\pmb a - \pmb b - k\pmb c = \pmb 0$, so one can take $x = k + 1, y = -1$ and $z = -k$.
Now let's assume that there are $x, y, z$ where $x + y +z = 0$ such that $x \pmb a + y\pmb b + z\pmb c = \pmb 0$. Since $x, y, z$ are not zeros all at once, at least one is non-zero. Let's assume it's $x \neq 0$. Due to the $x+y+z = 0$ at least one more number is non-zero. Let's assume it's $y \neq 0$
$x \pmb a + y \pmb b - (x + y) \pmb c = \pmb 0\\ (\pmb a - \pmb c) x + (\pmb b - \pmb c) y = \pmb 0$.
Last equation means that $\pmb a - \pmb c$ and $\pmb b - \pmb c$ are collinear. If one recalls that $\pmb b - \pmb a = \overrightarrow{\pmb {AB}}$ and $\pmb c - \pmb a = \overrightarrow{\pmb {AC}}$ then one can say that $A, B$ and $C$ are collinear as well.
2) Same thought can be used for coplanar case. With one more extra term.