Limit current to a load
3 × 1.2 Ω in series is 3.6 Ω.
\$ P = \frac {V^2}R = \frac {5^2}{3.6} = 6.9\ \text {W} \$.
And the current is \$ I = \frac V R = \frac 5 {3.6} = 1.38 \ \text A \$.
Your PSU is only capable of 5 W or 1 A so it's not big enough.
3 × 1.2 Ω in parallel is 0.4 Ω. I'll leave you to work out the current that will draw.