system of equations - when does it have a solution? Find all values of $t$ for which the system of equations 
$$\begin{array} 22x_1 + x_2 + 4x_3 + 3x_4 = 1\\
x_1 + 3x_2 + 2x_3 − x_4 = 3t\\
x_1 + x_2 + 2x_3 + x_4 = t^2 \end{array}$$
has a solution?
I was given a theorem, that system has a solution, when column vector of RHS lies in the subspace spanned by column vectors of LHS. If we take respective column vectors, and notice that third is a scalar multiple of the first column, we get three linearly independent vectors. What I don't get is, why should there be particular $t$'s, for which the system doesn't have a solution, as if we have three linearly independent (column) vectors, they should span $\mathbb{R^3}$, and thus we could find solution for any set of $t$'s. 
I suppose I'm wrong, but where's the mistake, and how should I check then, which $t$'s suffice? 
 A: Using Angela's observation, the system reduces to $$2x_1+x_2=1\\x_1+3x_2=3t\\x_1+x_2=t^2$$
Subtracting the third line from the first, we get $x_1=1-t^2$.  The second line then gives $3x_2=3t-(1-t^2)=t^2+3t-1$, while the third line gives $x_2=t^2-(1-t^2)=2t^2-1$.  Hence for the system to have a solution these must agree, i.e. $\frac{1}{3}(t^2+3t-1)=2t^2-1$ or $t^2+3t-1=6t^2-3$.  This is a quadratic equation $5t^2-3t-2=0$ with two solutions, and copper.hat kindly found them explicitly.
A: Let $A$ be the matrix of the left hand side. 
Notice that $A (-2, 0 ,1, 0)^T = 0$ and $A(0, -1, 1, -1)^T = 0$. Also, $ \operatorname{sp} \{ (2,1,1)^T, (1 -2, 0)^T\} \subset {\cal R} A $, hence using the rank nullity theorem, we have $\dim {\cal R} A = 2$.
Aside: Consider $f(t) = (1,3t,t^2)^T$. We have $f(0), f(1), f(2)$ are linearly independent, hence it is impossible to find solutions for all $t$.
So, you need to solve $f(t) \in {\cal R} A  = \operatorname{sp} \{ (2,1,1)^T, (1 -2, 0)^T\}$. Explicitly, find $x,y,t$ such that $f(t) = (1,3t,t^2)^T =x (2,1,1)^T + y(1 -2, 0)^T $.
We see that $x=t^2$, $y = \frac{1}{2} t (t-3)$, and the first equation then gives 
$2 t^2 + \frac{1}{2} t(t-3) = 1$, which simplifies to $5 t^2-3t-2 = 0$.
This gives $t = 1$ or $t=-\frac{2}{5}$.
Addendum: Let me eliminate matrix methods from the above. Note that the third column is $-2$ times the first. So we need only worry about one of these, I will pick the first $(2,1,1)^T$. Notice that the third column equals the second plus the fourth column, so we need only worry about one of these. However, for computational simplicity, I want a zero in the vector, so instead of using the second, I will use the first minus the second $(1 -2, 0)^T$. The range of possible right hand sides is then given by $x (2,1,1)^T + y(1 -2, 0)^T $, where $x,y$ range over the real numbers. The remainder of the argument above is 'matrix free'.
