Given set 1,2,3,5,6,7
Here sum is 1+2+3+5+6+7 =24
So there will be two cases
Case (i) When numbers are 1,2,5,6,7
Arrangement will be ____2&____6
In unit place 2 & 6 can take place because for multiple of 6 number should be even & sum of digits should be multiple of 3.
So total way in which remaining number can be taken will be 4!×2 =48 ways
Case (ii) When numbers are 1,2,3,5,7
Arrangement will be ____2
So total way in which remaining number can be taken will be 4!=24 ways
Adding case (i) & case (ii), we get
required 5-digit numbers =24+48=72