Condition for infinitely many solutions

Question

$$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ For how many values of $\alpha$ does this system of equations have infinitely many solutions ?
This was the question I have to solve.
Now I know since 3rd column is scalar multiple of 2nd column, so column space $C(A)$ of coefficient matrix $A$ is a 2D plane in 3D space.
So there is only one value of $\alpha$, which will put this vector $b$ into that 2D plane. Now my question is do we have infintely many solutions then, I mean I can't understand how there are multiple linear combinations of 3 column vectors of $A$ such that it results into same vector $b$ ?

Answer 1

The point is that the system has a kernel; since the column space is not 3-dimensional there are nontrivial linear combinations of the columns that vanish, namely multiples of $x=0,y=4,z=1$. So given one solution, you find another by adding any element of this kernel. Do be sure to check your claim about there being exactly one $\alpha,$ though: not every line intersects every plane, and a given line intersects some planes infinitely often.