Weird Kirchhoffs laws circuit
The first thing I usually do is redraw such schematics so that they make a little more sense.



simulate this circuit &ndash; Schematic created using CircuitLab
You can analyze this in a number of ways. My preference is nodal analysis, which depends more on KCL. Mesh analysis is more about KVL. Both work. The above schematic has been prepared for either method.
Nodal
$$\begin{align*}
\frac{V_A}{R_3}+\frac{V_A}{R_4}+\frac{V_A}{R_5}&amp;=\frac{V_B}{R_3}+\frac{V_2}{R_4}+\frac{V_D}{R_5}\\\\
\frac{V_B}{R_1}+\frac{V_B}{R_2}+\frac{V_B}{R_3}&amp;=\frac{V_C}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_A}{R_3}\\\\
\frac{V_D}{R_5}&amp;=I_1+\frac{V_A}{R_5}\\\\
\frac{V_C}{R_1}+I_1&amp;=\frac{V_B}{R_1}\\\\
V_D&amp;=V_C+V_1
\end{align*}$$
The above analysis provides you with all the voltages and one current. But with the voltages you can work out the rest.
Mesh
$$\begin{align*}
0\:\text{V}+V_2-I_2\cdot R_4-\left(I_1+I_2\right)\cdot R_3-I_2\cdot R_2&amp;=0\:\text{V}\\\\
V_C+V_1-I_1\cdot R_5-\left(I_1+I_2\right)\cdot R_3-I_1\cdot R_1&amp;=V_C
\end{align*}$$
The above analysis provides you with the two indicated currents. You can work out the rest from there.
Summary
You can easily see that the current through \$R_3\$ is the sum of \$I_1\$ and \$I_2\$. It's not something you need to directly solve for, since it falls out once you know the other two currents.
Since there is no possible loop for \$I_3\$, you can't use it to develop a new loop equation. It's an outcome of the other two loops.