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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)\$ ?

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

enter image description here

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A=12V put that in your equations and solve for the voltage at the other node (B and C).

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Hint: what is \$V_a\$?

schematic

simulate this circuit – Schematic created using CircuitLab

Also, your equation A is incorrect; this assumes the current flowing through the 12V source is 0. Luckily, you don't need this equation because you already have 2 equations and 2 unknowns.

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