That's clever. The 2nd relation is $x+ y+ z= \sqrt{x+ y+ z- 3}+ 15$. Subtracting 3 from both sides, $x+ y+ z- 3= \sqrt{x+ y+ z- 3}+ 13$. Letting u= x+ y+ z- 3, We can write that as $u= \sqrt{u}+ 12$. And that is the same as $\sqrt{u}= u- 12$. Squaring both sides, $u= (u- 12)^2= u^2- 24u+ 144$ s that $u^2- 25u+ 144= (u- 16)(u- 9)= 0$. The roots of that are u= 16 and u= 9. If u= 16 then x+ y+ z= u+ 3= 19. If u= 9 the x+ y+ z= u+ 3= 12.
Having squared both sides we need to check for spurious solutions that might have been introduced. If x+ y+ z= 19, then the equation becomes $19= \sqrt{19- 3}+15= 4+ 15$ which is correct. If x+ y+ z= 12, then the equation becomes $12= \sqrt{12- 3}+ 15= 3+ 15$ which is not correct (we got this "spurious solution" because -3+ 15 is equal to 12.)
I'm impressed!
from (1) and above
$ x = 4, y = 3, z = 12 $ giving $x-y+z = 13$
or
$x = 8, y = 6, z = 5$ giving $x-y+z = 7$