So this is what your run-off-the-mill 3-phase heater looks like

simulate this circuit – Schematic created using CircuitLab
$$R_1=R_2=R_3=R$$
Notice that at the center point, your neutral wire could be connected — but due to the 120° phase offset summing to 360°, that point is at a constant 0V, anyways, so this is, strictly speaking, optional¹.
Now, apply the Y-Δ transform:

simulate this circuit
For symmetry reasons, we can deduct that \$R_{12}=R_{23}=R_{31}=R_\Lambda\$.
$$R_\Lambda = \frac{RR+RR+RR}{R}= \frac{3R^2}{R} = 3R$$
The important part here is that *we can't tell whether it's Y or Δ from the outside**! In other words, if your circuit breaker trips for \$R\$ in Y-configuration, it'll trip for \$3R\$ in Δ-configuration.
Since the power going into the two circuits can't be different, it's impossible that you find one to give you more heating than the other.
¹ I was actually at a friend's place where one of the three phases in her multi-apartment building was dead – so she had no light in the kitchen. Unless she turned on the heater – which led to the ungrounded center point of the Y-configured heater getting shifted, and thus, the dead P3 getting seeing enough voltage to turn on an incandescent light bulb dimly. Scary stuff.