I₁ = 1 A; I₂ = -2 A; I₃ = 1 A

There are a couple of ways to check whether the result is correct. One is to label the potential at different points in the circuit and make sure everything is consistent. Choose an arbitrary point to be zero (I chose the negative side of battery 1) and label potentials relative to that.

Going across a battery from - to + raises the potential by the battery voltage. Going from + to - drops it.

The potential difference across a resistor is IR. The current flows through a resistor from higher potential to lower potential.

The fact that I₂ is negative means that the direction we chose for I₂ was wrong. Make sure you remember that that current really goes the other way.

Another way to check your answer is to make sure that the power everywhere in the circuit is consistent with the Law of Conservation of Energy.

Using I²R, the power dissipated in each resistor is:

For R₁: P = (1)²1 = 1 W
For R₂: P = (2)²2 = 8 W
For R₃: P = (1)²3 = 3 W
For R₄: P = (1)²4 = 4 W

That adds up to 16 W dissipated in the resistors.

The power associated with the batteries can be determined using P = eI.

For e₁: P₁ = 12*1 = 12 W
For e₂: P₂ = 3*2 = 6 W
For e₃: P₃ = 10*1 = 10 W

At first glance it looks like the batteries are putting more power into the circuit than is being dissipated. However, battery e₂ has current running through it from + to -, so it is being re-charged and is taking energy out of the circuit.

Batteries e₁ and e₃ supply 22 W of power to the circuit.
The resistors plus battery e₂ use 22 W of power.

Everything adds up, as it should.