Contradiction while solving circuit problem!
Let's set up the mesh equations as you have the schematic drawn up:

$$\begin{align*}
0\:\text{V}-200\cdot I_3-600\cdot I_3-400\cdot\left(I_3-I_2\right)-V_{5\:\text{mA}}&amp;= 0\:\text{V}\\\\
0\:\text{V}-R_V\cdot\left(I_2-I_1\right)-400\cdot\left(I_2-I_3\right)-V_{V_2\over 400}&amp;=0\:\text{V}\\\\
0\:\text{V}+4\:\text{V}-500\cdot I_1+V_{5\:\text{mA}}-R_V\cdot\left(I_1-I_2\right)&amp;=0\:\text{V}\\\\
I_1-I_3&amp;=5\:\text{mA}\\\\
I_2=\frac{V_2=200\cdot I_3}{400}&amp;=\frac{I_3}{2}
\end{align*}$$

This provides 5 equations and five unknowns: \$I_1\$, \$I_2\$, \$I_3\$, \$V_{5\:\text{mA}}\$, \$V_{V_2\over 400}\$.

Note that this includes the fact that there are, in fact, voltage drops across your current sources. Those are just two more variables, as shown.

Solving this yields 5 functions which depend upon \$R_V\$.



I can't tell you if \$I_2=I_3=0\:\text{A}\$ without knowing more about \$R_V\$. And I don't understand \$.5\:N\:k\$. (Does it mean \$500\cdot N\$?) So I'm stuck at this point, if you are looking for numerical results that aren't functions of \$R_V\$ (or N.)



(You can get \$I_2=I_3=0\:\text{A}\$ if and only if \$R_V=300\:\Omega\$. For obvious reasons.)