If 120 J of thermal energy is incident on area $3 \mathrm{~m}^2$, the amount of heat transmitted is 12 J, coefficient of absorption is $0.6$, then the amount of heat reflected is

The total incident thermal energy $Q_{\text{incident}} = 120\text{ J}$ follows the conservation principle: $Q_{\text{incident}} = Q_{\text{absorbed}} + Q_{\text{reflected}} + Q_{\text{transmitted}}$ Given the absorption coefficient $a = 0.6$ and transmitted energy $Q_{\text{transmitted}} = 12\text{ J}$, the absorbed energy is: $Q_{\text{absorbed}} = a \times Q_{\text{incident}} = 0.6 \times 120 = 72\text{ J}$ Substituting into the conservation equation: $120 = 72 + Q_{\text{reflected}} + 12$ $Q_{\text{reflected}} = 120 - (72 + 12) = 36\text{ J}$

Final answer: $\boxed{36\text{ J}}$