Why am I getting negative current on the right loop of the Kirchhoff's circuit? 
I have the above circuit and I calculated that the:
left loop = I1(Rc+4) - I2(Rc)=12

right loop = I1(Rc) - I2(Rc+3) = 5
I then generated numbers from 0 to 5 ohms with stepsize 0.1.
After replacing these values with the left loop to calculate the current on the left loop, I got all posative current values, but when I put it in the right loop equations, some of my current numbers turned to be negative. Does it mean the direction of the flow is wrong in the right equation? or is it due to the negative voltage on the right?
For example when I have Rc=0, then current on the right is: -1.6667
This is my calculations below:

Do I keep the values obtained negative if I want to graph current vs resistance?
 A: When you get negative current, it does not mean it is wrong. It only means that the actual current direction is in the opposite way to the assumed direction.

It is not clear from your schematic what is polarity of the voltage source on the right. If it is minus $5 \text{V}$ then your equation for the right-hand side loop has wrong polarity for the selected current direction (clockwise for both currents). The correct equations are
$$
\left\{
\begin{array}{ll}
I_1 (R_c + 4) - I_2 R_c = 12 \\
I_1 R_c - I_2 (R_c + 3) = -5
\end{array}
\right.
$$
which results in
$$I_1 = \frac{36 + 17 R_c}{12 + 7 R_c} \qquad \text{and} \qquad I_2 = \frac{20 + 17 R_c}{12 + 7 R_c}$$
When you get a result like this, you should always test it against some extreme (limiting) cases:

*

*For $R_c = 0$ you have a short-circuit in the middle, which means two loops are completely independent. The two currents are $I_1 = 12 / (3 + 1) = 3 \text{ A}$ and $I_2 = 5 / (1 + 2) = 5/3 \text{ A}$.

*For $R_c = \infty$ you have an open-circuit in the middle, which means there is only one loop with two voltage sources. The two currents are $I_1 = I_2 = (12 + 5) / (3 + 2 + 1 + 1) = \frac{17}{7} \text{ A}$.

A: Like @Marko Gulin said, a negative number doesn't mean its wrong, it just means the direction of the loop current you chose is opposite to the actual direction of the loop current.
More importantly, you should show the direction of the loop currents that your equations are based on so that we don't have to take the time to figure out if your loop equations are correct.
Hope this helps.
A: In the limit $R_\text{C}\to\infty$, this becomes a single-loop circuit where the 12V source pushes current the “wrong way” through the 5V source. This is one type of battery charger.
In the limit $R_\text{C}\to 0$, you have two independent circuits which are both sending current through the same conducting segment.  This is also a common configuration for real circuits; we frequently call the shared segment “the ground.”
Between your electronic diagram, your handwritten notes, and your equation, I can’t tell whether the constant voltage on the right is $V_2 = \rm 5\,V$, or $V_2 = \rm -5\,V$, or $V_2 = \rm 0.5\,V$.  If it’s actually negative, and you’ve written a negative value instead of putting the battery symbol’s long lines on the bottom, you are suffering for no good reason.  If it’s positive, the equation in your question (v3) contains at least one sign error.
