Preface note: I originally used \$3\:\text{A}\$ as the current source. But I mis-read the problem, as stated. They specify, on the schematic, that the out-flowing current source value is \$-3\:\text{A}\$. The original error is mine. It is fixed, now, below.
The node is (by super-position):
$$\overbrace{\frac{V_2}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2}{7\:\Omega}+\left(-3\:\text{A}\right)}^{\text{Outward-flowing currents}}=\overbrace{\frac{V_1}{3\:\Omega}+\frac{0\:\text{V}}{1\:\Omega}+\frac{V_3}{7\:\Omega}}^{\text{Inward-flowing currents}}$$
Moving everything from the right side over to the left side gives:
$$\frac{V_2-V_1}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2-V_3}{7\:\Omega}+\left(-3\:\text{A}\right)=0\:\text{A}$$
Moving the constant to the right and you have:
$$\frac{V_2-V_1}{3\:\Omega}+\frac{V_2}{1\:\Omega}+\frac{V_2-V_3}{7\:\Omega}=3\:\text{A}$$
Does that help at all?
Note
Voltage spills away from a node via available paths and voltage spills into a node from nearby nodes via the same available paths. The only caveat is that a current source can only flow into, or flow out of, a node. But not both directions. (Current sources only work one way. So it will either appear on the outward-flowing side or on the inward-flowing side -- but not both sides.)
