Mesh Analysis on CCT with Voltage and Current Source
Here's the schematic, redrawn slightly to my own taste:







simulate this circuit &ndash; Schematic created using CircuitLab

Starting at the lower right-hand corner of each loop:

$$\begin{align*}
0\:\text{V}+12\:\text{V}-R_3\left(I_1-I_2\right)-V_{I_1}-R_4\left(I_1-I_3\right) &amp;= 0\:\text{V}\tag{$I_1$}\\\\
0\:\text{V}+V_{I_1}-R_3\left(I_2-I_1\right)-R_1\: I_2 -R_5\left(I_2-I_3\right) &amp;= 0\:\text{V}\tag{$I_2$}\\\\
0\:\text{V}-R_4\left(I_3-I_1\right)-R_5\left(I_3-I_2\right)-R_2\: I_3 &amp;= 0\:\text{V}\tag{$I_3$}\\\\
I_0=I_1-I_2&amp;=6\:\text{mA}\tag{Known}
\end{align*}$$

If you look closely, the variables you need to solve for are \$I_1\$, \$I_2\$, \$I_3\$, and \$V_{I_1}\$. You have four equations and four unknowns.

No supermesh stuff, either.

Once you solve for those values, just work out the current in \$R_5\$. From that, you can figure out the voltage drop magnitude and also the polarity (from the current direction.)



Using Sympy/Sage, it's just:

 i1, i2, i3, vi0 = S('i1 i2 i3 vi0'.split())
 e1 = Eq( 0 + 12 - 4E3*(i1-i2) - vi0 - 2E3*(i1-i3), 0 )
 e2 = Eq( 0 + vi0 - 4E3*(i2-i1) - 8E3*i2 - 6E3*(i2-i3), 0 )
 e3 = Eq( 0 - 2E3*(i3-i1) - 6E3*(i3-i2) - 8E3*i3, 0 )
 e4 = Eq( i1 - i2, 6E-3 )
 r = solve( [e1, e2, e3, e4], [i1, i2, i3, vi0] )
 r