Answer
$a.\quad 24$
$b.\quad\{(x,y,z)|\quad x+y+z\leq 10\},$
which contains points on the plane x+y+z=10,
or in the region (below the plane) containing the origin. (see image)
Work Step by Step
$a.$
$g(1,2,3)=1^{3}\cdot 2^{2}\cdot 3\sqrt{10-1-2-3}=12\sqrt{4}=24$
$b.$
Because of the restrictions for square roots, g is defined when
$10-x-y-z \geq 0$
$10\geq x+y+z$
$\mathbb{R}^{3}$ is divided into two regions by the plane $x+y+z=10$.
The domain includes the plane.
Testing (0,0,0) the inequality is satisfied, so the domain includes the region containing the origin.
