Find an ordered triple (x,y,z) of real numbers

Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrtx + sqrty +sqrtz = 10
x+y+z=42
(sqrtxyz) = 20

Hello Guest!

$\sqrt{x}+\sqrt{y}+\sqrt{z}=10$

$x+y+z=42$

$\sqrt{xyz}=20$

$xyz=400$

$z=\frac{400}{xy}=42-x-y$

$z= (10-\sqrt{x}-\sqrt{y})^2$

$I.\ \frac{400}{xy}=42-x-y$

$II.\ \frac{400}{xy}=(10-\sqrt{x}-\sqrt{y})^2$

$y\in\{-3,3\}\\ x\in \{-2,2\}$

$z\in \{5,9,11,15\}$

$z=3$

$x+y+3=42\\ x+y=39$

Error! To be continued.

  !

As follows:

My statement above that "There are only four possibilities ..." assumes that we are dealing with integers only.