You need to calculate the junction temperature when dissipating 10.5 watts at an ambient of 70 °C. The 10.5 watts comes from the fact that the linear regulator is dropping 7 volts to convert 12 volts to 5 volts. That volt-drop × current (1.5 amps) equals 10.5 watts.
You calculate junction temperature based on the data sheet for the regulator and your heat-sink specification whilst taking into account the minimum air-flow needed for the heat-sink to be effective. You might also need to account for local ambient warming by a few degrees.
So, get the figure for junction to case thermal resistance (\$\theta_{JC}\$) from the data sheet and multiply that by power dissipated. The result is the temperature that the junction will be at when the case is held at some ambient temperature. So, if the number realized is (say) 60 °C and your ambient is 70 °C then the junction will be at 130 °C.
Then you need to factor in the heat-sink. It might have a thermal-resistance to ambient of 2 °C per watt and that means that the 10.5 watts flowing through it will raise its local temperature by 21 °C.
You then add this to your local ambient temperature of 70 °C to get 91 °C and then your junction will be at 151 °C and this might be too much but, only your data sheet will tell you.
There are a few more nuances you might need to know but I'll leave that for now.
Will this damage the IC over time or reduce the reliability of
regulator IC?
If the normal expected maximum junction temperature is 150 °C and you continuously run close to or above this temperature, you will reduce the lifetime and reliability of the device.
