Nodal Analysis current source confusion

My instructor told me to "kill" sources (treating current sources as open and voltage sources as short). From this my third node would have to be in the top right corner.

Suppose its called A. Would the following be the correct KCL for it?

$$A: \frac{A-C}{2k} - (4mA) = 0$$

$$B: \frac{B-C}{6k} + \frac{B-Ref}{3k} + (4mA) = 0$$

$$C: \frac{C-A}{2k} + \frac{C-B}{6k} + \frac{C-Ref}{4k} = 0$$

What I am confused on is where the 12V comes in to play. Does it just get added to A (i.e $A=(A+12V)$ ?

$$A:\frac{A+12V-C}{2K} - (4mA) = 0$$

Hint: what is $V_a$?