In the above circuit the current in each resistance is:

Kirchoff's voltage law in loop 1 : 6+2+(i1−i2)1=0 ⇒i1−i2=0 or i1=i2—(1)
Kirchoff's Voltage law in loop 2 : 6+(i2−i1)1+2+(i2−i3)1=0 ⇒i2−i3=0 or i2=i3—(2)
Kirchoff's Voltage law in loop 3 : 6+(i3−i2)1+2+(i3−i1)1=0 ⇒i3=0
From (2) we get i2=i3=0
From (1) we get i1=i2=0 ⇒i1=i2=i3=0