Is an arbitrary number of the form xyzxyz divisible by 7, 11, 13?

Hint:

$$7\cdot11\cdot13=1001$$

Every number of that form is divisible by $7$, $11$ and $13$:

$$\underbrace{xyz}_\text{1000xyz}~xyz = 1000xyz + xyz = 1001xyz = 7\cdot11\cdot13\cdot xyz$$

Here is a more naive approach. Just using the usual criteria.

Divisibility by 7: $$z+3y+2x+6z+4y+5x=7z+7y+7x=7 (x+y+z), $$so $xyzxyz $ is a multiple of $7$.

Divisibility by $11$: $$z-y+x-z+y-x=0, $$ so $xyzxyz $ is a multiple of $11$.

Divisibility by $13$: $$z-3y-4x-z+3y+4x=0, $$ so $xyzxyz $ is a multiple of $13$.