If the given number is $xyz$ then using this number to form two more 3-digit numbers, like:

The first number is $zxy$, and the other number is $yzx$.

These can also be written as

$xyz=100x+10y+z$

$zxy=100z+10x+y$

$yzx=100y+10z+x$

Summing these three, we get:

$111(x+y+z)$ which is always divisible by $37$.