"Show that $(a \ \ 3)$ and $(2 \ \ \ 1-a)$ are linearly independent vectors for each value of a"
This means that given scalars $C1,C2$ their scalar sum should equal $0$:
$aC1 + 2C2 = 0$
$3C1 + C2-aC2 = 0$
$\Rightarrow 6C1 - aC1 + a^2C1 = 0.$
Don't know what I need to do now or if this is wrong already.
Your approach is workable. As for the comment where you got $\Rightarrow$ from one solves the first equation for $C_2 = -aC_1/2$ and inserts that in the second equation. What you're basically doing here is a proof by contradiction - you assume that there is a (non-trivial) linear combination of the vectors that's $0$.
The requirement is also $C_1$ and $C_2$ should not both be zero. You know that if one is zero the other must be zero for the sum to be zero. So you basically know that this requires $C_1\ne 0$. So you can divide both sides with $C_1$ and get the equation:
$$\left(a-{1\over2}\right)^2 + {23\over4} = \left(a-{1\over2}\right)^2 - {1\over4} + 6 = a^2 - a + 6 = 0$$
Which isn't possible with real $a$ since $(a-1/2)^2\ge 0$ and $23/4>0$ so the LHS (left hand side) is certainly greater than zero and not equal to zero. Now as we see that the equation is certainly false for all $a$ we have a contradiction an we can conclude that our assumption that theres a (non-trivial) linear combination is false.
That is whatever value of $a$ we have no (non-trivial) linear combination of the vectors that's the null vector.