Impedance of the circuit between given points
For case #1, we have:







simulate this circuit &ndash; Schematic created using CircuitLab

Applying nodal analysis:

$$\begin{align*}
\frac{V_a}{R_4}+\frac{V_a}{R_6}&amp;=1\:\text{A}+\frac{V_b}{R_4}+\frac{V_c}{R_6}\\\\
\frac{V_b}{R_2}+\frac{V_b}{R_4}+\frac{V_b}{R_5}&amp;=\frac{V_d}{R_2}+\frac{V_a}{R_4}+\frac{V_c}{R_5}\\\\
\frac{V_c}{R_3}+\frac{V_c}{R_5}+\frac{V_c}{R_6}&amp;=\frac{V_d}{R_3}+\frac{V_b}{R_5}+\frac{V_a}{R_6}\\\\
\frac{V_d}{R_1}+\frac{V_d}{R_2}+\frac{V_d}{R_3}&amp;=\frac{V_b}{R_2}+\frac{V_c}{R_3}
\end{align*}$$

From which we find \$V_a=9.46715328467153\:\text{V}\$ and therefore \$R_\text{TH}\approx 9.47\:\Omega\$.

Moving the current source to the next position, we get case #2:







simulate this circuit

We find that:

$$\begin{align*}
\frac{V_a}{R_1}+\frac{V_a}{R_4}+\frac{V_a}{R_6}&amp;=\frac{V_d}{R_1}+\frac{V_b}{R_4}+\frac{V_c}{R_6}\\\\
\frac{V_b}{R_4}+\frac{V_b}{R_5}&amp;=1\:\text{A}+\frac{V_a}{R_4}+\frac{V_c}{R_5}\\\\
\frac{V_c}{R_3}+\frac{V_c}{R_5}+\frac{V_c}{R_6}&amp;=\frac{V_d}{R_3}+\frac{V_b}{R_5}+\frac{V_a}{R_6}\\\\
\frac{V_d}{R_1}+\frac{V_d}{R_2}+\frac{V_d}{R_3}&amp;=\frac{V_a}{R_1}+\frac{V_c}{R_3}
\end{align*}$$

From which we find \$V_b=10.6311475409836\:\text{V}\$ and therefore \$R_\text{TH}\approx 10.63\:\Omega\$.

Moving the current source to the next position, we get case #3:







simulate this circuit

We find that:

$$\begin{align*}
\frac{V_a}{R_1}+\frac{V_a}{R_4}+\frac{V_a}{R_6}&amp;=\frac{V_d}{R_1}+\frac{V_b}{R_4}+\frac{V_c}{R_6}\\\\
\frac{V_b}{R_2}+\frac{V_b}{R_4}+\frac{V_b}{R_5}&amp;=\frac{V_d}{R_2}+\frac{V_a}{R_4}\\\\
\frac{V_c}{R_3}+\frac{V_c}{R_6}&amp;=1\:\text{A}+\frac{V_d}{R_3}+\frac{V_a}{R_6}\\\\
\frac{V_d}{R_1}+\frac{V_d}{R_2}+\frac{V_d}{R_3}&amp;=\frac{V_a}{R_1}+\frac{V_b}{R_2}+\frac{V_c}{R_3}
\end{align*}$$

From which we find \$V_c=7.58479532163743\:\text{V}\$ and therefore \$R_\text{TH}\approx 7.58\:\Omega\$.