Kirchoff law in AC circuit
No, that's not correct. You say that the voltage across the resistor is 12j V, but that assumes that all of the 3j A go through the resistor, and there's also the path through the voltage source and inductor.

I would call the lower connection of the resistor ground, and the upper V1. Now try to find V1 by applying KCL to the node (that's the sum of three currents).  

edit (re your update of the question)
Your arrows suggest that you want to want to find the currents in the two loops, but that's not how KCL works. This doesn't cause anything but problems: I1 seems to be -I2 if I look at how they flow through the resistor, and I3 = I2.

KCL: you pick a node: V1. Then assign a current name to all branches that start from V1: I1 is the current source at the left, I2 is the current from V1 through the resistor to ground, and I3 is the current from V1 through the inductor and the voltage source to ground. Then I1 + I2 + I3 = 0. I1 is easy, that's given: 3j A. I2, also easy, a voltage difference over a resistor, write the equation for I2. For I3 you have two voltage sources, V1 and the -6j V source, and one impedance: the inductor. Again, write down I3 as a difference between two voltage sources divided by an impedance. 

Now you have three expressions for the currents, set their sum equal to zero, and you have an equation with just one variable: V1. Calculate V1, and fill its value in the equations for I2 and I3, and you're done.