Nodal question with current-controlled voltage source connected to independent voltage source

so i've got a question here

the question is to calculate v1 & iΦ, of course for nodal analysis (v1/24) + ((v1-(60+6iΦ))/3) + ((v1-va)/2) = 0, but (with va as the node left of 2 ohm resistor) how to calculate va? is it 60+6iΦ? 60-6iΦ? 60? or is it something else?

Answer

The first thing to do is to redraw the schematic into a somewhat better (more readable) layout. (You can read a short discussion I wrote here.)

^{simulate this circuit – Schematic created using CircuitLab}

(Note that you get to assign exactly one node to the value \$0\:\text{V}\$. You can only do that once, to one node. No more. There are convenient places to make such assignments. But no bright line reasons to choose one over another. You can move the ground assignment to anywhere you please and the basic analysis results will be the same. The absolute values will be different. But the relative values won't change, no matter how you assign ground.)

From the above, you can easily see that:

$$\begin{align*} \frac{V_A}{R_1}+\frac{V_A}{R_2}+\frac{V_A}{R_3} &= \frac{60\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_B}{R_3}\\\\ V_B &= 60\:\text{V}+6\cdot I_0\\\\ I_0 &= \frac{V_B-V_A}{R_3} \end{align*}$$

The above is readily solvable for \$V_A\$, \$V_B\$, and \$I_0\$.