#### Answer

$D=$ {$(x,y,z) | x^2+y^2+z^2\leq 1$}

#### Work Step by Step

As we are given that $f(x,y,z)=arcsin(x^2+y^2+z^2)$
The function $f(x,y,z)=arcsin(x^2+y^2+z^2)$ represents a trigonometric function which cannot be negative and exists only for positive numbers.
Also, $\sin x$ function lies in between -1 and +1.
Thus,
$ -1 \leq x^2+y^2+z^2\leq 1$
or, $x^2+y^2+z^2\neq -1 $
This means that $x^2+y^2+z^2\leq 1$
Hence, Domain: $D=$ {$(x,y,z) | x^2+y^2+z^2\leq 1$}