Definition: A set of vectors $\{\vec v_1,\ \vec v_2,\ \vec v_3,\ \dots,\ \vec v_k\}$ is linearly independent if the ONLY solution to the equation $a_1 \vec v_1 + a_2 \vec v_2 + \cdots + a_k \vec v_k = \vec 0$ is $a_1 =0,\ a_2 = 0,\ \dots, a_k=0$.
Another way to say this is that no vector in the set $\{\vec v_1,\ \vec v_2,\ \vec v_3,\ \dots,\ \vec v_k\}$ can be found by a linear combination of the others (adding any combination of the others with any coefficients).
Defintion: A set of vectors $\{\vec v_1,\ \vec v_2,\ \vec v_3,\ \dots,\ \vec v_k\}$ is linearly dependent if it is NOT linearly independent.
That is, there exists at least one solution to the equation $a_1 \vec v_1 + a_2 \vec v_2 + \cdots + a_k \vec v_k = \vec 0$ where NOT EVERY $a_i$ is $0$.
Of course, given a linearly DEPENDENT set $\{\vec v_1,\ \vec v_2,\ \vec v_3,\ \dots,\ \vec v_k\}$, the equation $a_1 \vec v_1 + a_2 \vec v_2 + \cdots + a_k \vec v_k = \vec 0$ is STILL TRUE when $a_1 =0,\ a_2 = 0,\ \dots, a_k=0$. But the point is that it is NOT THE ONLY SOLUTION. In fact, it is called the "trivial solution".
What your question is asking you to do is find some coefficients $a_1,\ a_2, a_3$, not all of which are $0$, such that the equation $a_1 (3,2) + a_2(-1,1) + a_3(4,0) = (0,0)$.
Notice that this question is IDENTICAL to solving the system of equations: $\begin{cases}3a_1 -a_2 + 4a_3=0 \\ 2a_1 + a_2 =0 \end{cases}$ for the variables $a_1,\ a_2,\ a_3$. You can then clearly see that $a_1=0,\ a_2=0,\ a_3=0$ IS a solution, but your problem specifically tells you to find a DIFFERENT one. (Thus why you finding $c_3=0,\ c_1=\frac {-4}5 c_3 = 0,\ c_2=\frac 85 c_3 =0$ is NOT what you're supposed to be looking for.)
This system of equations (AKA vector equation) has infinitely many solutions for $a_1,\ a_2, a_3$, you just need to find one that is "non-trivial". The are $2$ simple methods (and a whole bunch of slightly less simple ones) that you could use to do this:
Method #$1$: Because this system is so small, you could just try setting one of the variables equal to an arbitrary value and solving for the other two. For instance, try solving this system where $a_1=1$ (Notice I did NOT set it equal to $0$). Can you find values of $a_2$ and $a_3$ that then solve this system?
Method #$2$: You could use an augmented matrix:
$$\left[\begin{array}{ccc|c} 3 & -1 & 4 & 0 \\ 2 & 1 & 0 & 0 \end{array}\right]$$
What you want to do here is use Gauss-Jordan elimination to find the reduced row echelon form. Then you'll see that there is at least one column without a leading $1$. Then your first column represents $a_1$, your second $a_2$, and 3rd is $a_3$. Whichever column(s) doesn't have a leading $1$, you can set that variable equation to an arbitrary value. For instance, if you matrix ends up looking like:
$$\left[\begin{array}{ccc|c} 1 & 0 & 3 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right]$$
then you'll want to set $a_3 = t$ for some arbitrary $t$. Then solve for $a_1$ and $a_2$ in terms of $t$. For the matrix I just came up with, you'd have $a_1=0-3t=-3t$, $a_2=0-t=-t$ and $a_3=t$. Whatever you get at this point will be your GENERAL SOLUTION.
Then to find any SPECIFIC solution, set your $t$ (or whatever this arbitrary value) equal to ANY NONZERO VALUE. For instance in my example let $t=1$. Then one of the infinite number of solutions to this is $a_1=-3(1)=-3$, $a_2=-(1)=-1$ and $a_3=1$.
Best Answer
It is possible that they can be linearly dependent. Consider the map $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which maps $(x,y)$ to $(x,0)$. Then $(1,0)$ and $(1,1)$ are linearly independent, but $T(1,0) = T(1,1) = (1,0)$.
The problem with your argument is that $T(c_1 u + c_2 v) = T(0) = 0$ does not imply that $c_1 u + c_2 v = 0$. It just means that $c_1 u + c_ 2 v$ is a solution of $T(x) = 0$. When $u$ and $v$ are linearly independent, it must be the case that $c_1 u + c_2 v$ is nonzero, so this solution is non-trivial.