Maybe you mean to ask,
Does the point $(2,1)$ fall on the line parametrized by $a(x),b(x)$ or $c(x)$ if any?
$$ a(x) = (2,1)=(1,x) \ \ \Rightarrow \ \ 2=1, 1=x $$
$$ b(x) = (2,1)=(x,1) \ \ \Rightarrow \ \ 2=x, 1=1 $$
$$ c(x) = (2,1)=(x,2x) \ \ \Rightarrow \ \ 2=x, 1=2x $$
Obviously $2=1$ is hard to solve, the equation for $b(x)$ shows $b(2)=(2,1)$ and the equation for $c(x)$ gives $x=2$ and $x=1/2$ which is a contradiction. Only $b$ contains $(2,1)$ in its image.
Following Bill's suggestion: you could find $c_1,c_2,c_3$ such that
$$ (2,1) = c_1(1,x)+c_2(x,1)+c_3(x,2x) $$
This equation has to hold for all $x$ if $(2,1)$ is in the span of $a,b,c$ as vector-valued functions.