There are several mistakes. But you still did pretty good. Let's look at the schematic:
simulate this circuit – Schematic created using CircuitLab
Now all of the important nodes have been labeled and a new unknown current identified.
Your first equation should be:
$$\begin{align*}
\frac{V_a}{R_1}+\frac{V_a}{R_2}+\frac{V_a}{R_3}&=\frac{V_1}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_b}{R_3}+I_1+I_2\\\\\therefore\\\\\frac{V_a-V_1}{R_1}+\frac{V_a-0\:\text{V}}{R_2}+\frac{V_a-V_b}{R_3}-I_1-I_2&=0\:\text{A}
\end{align*}$$
And you got that one right! So there's no problem, yet.
Your second equation should be:
$$\begin{align*}
\frac{V_b}{R_3}+\frac{V_b}{R_4}+\frac{V_b}{R_5}+I_1&=\frac{V_a}{R_3}+\frac{V_c}{R_4}+\frac{V_c+V_2}{R_5}+I_3\\\\\therefore\\\\\frac{V_b-V_a}{R_3}+\frac{V_b-V_c}{R_4}+\frac{V_b-V_c-V_2}{R_5}+I_1-I_3&=0\:\text{A}
\end{align*}$$
Note that this is quite different from yours. One of the differences is that you used \$22\:\Omega\$ instead of \$R_4=12\:\Omega\$, I think. But actually there are more serious errors, as you can see. So this is really minor, by comparison.
You need two more equations. One of them is very simple:
$$\begin{align*}
\frac{V_c+V_2}{R_5}&=\frac{V_b}{R_5}+I_4\\\\\therefore\\\\\frac{V_c+V_2-V_b}{R_5}-I_4&=0\:\text{A}
\end{align*}$$
The final one is:
$$\begin{align*}
\frac{V_c}{R_4}+\frac{V_c}{R_6}+I_4&=\frac{V_b}{R_4}+\frac{0\:\text{V}}{R_6}\\\\\therefore\\\\\frac{V_c-V_b}{R_4}+\frac{V_c-0\:\text{V}}{R_6}+I_4&=0\:\text{A}
\end{align*}$$
That's it. From these four you can work out all of \$V_a\$, \$V_b\$, \$V_c\$, and \$I_4\$.
Took me less than 10 minutes to write this out.
