Can i still calculate the \$I_3\$ without transforming norton to thevenin circuit?
You need nodal or mesh equations which is basically KCL or KVL applied a grand scale. Note the plural EQUATIONS, not singular. They are more systematic than whatever it is you are trying to do by applying KCL piecemeal and trying to get away with just one equation.

\$6+2=I_2+I_3 =&gt;8=I_2+I_3\$

You are just going to ignore that I_1 exists at that 4-way node and not I_3? This is because you are trying to use just one equation and shortcut to I_3. I_3 splits up into I_1 and I1 so by using I_3 and I1 in the same node equation you are counting I_3 twice while also ignoring I_1. By the way, I1 and I_1 are terrible naming schemes.

Try an equation for every 4 way node and 3 way node. Three equations for three unknowns I_1, I_2, I_3. Then solve simultaneously. When everything affects everything else you cannot solve for a small part without solving for everything.