#### Answer

(a) $g(1,2,3)=24$.
(b) The domain is $\mathcal{D}=\{(x,y,z)|x+y+z\leq10\}$.
Geometrically, it is the space under the plane (including it) that intercepts coordinate axes at $x=10$, $y=10$ and $z=10$, respectively.

#### Work Step by Step

(a) By direct substitution we have
$$g(1,2,3)=1^3\cdot2^2\cdot3\sqrt{10-1-2-3}=1\cdot4\cdot3\sqrt{4}=24.$$
(b) The argument of the square root has to be nonnegative i.e. $10-x-y-z\geq0$ which gives
$$x+y+z\le10.$$
Consider the equation $x+y+z=10$. This is the equation of the plane determined by the triangle with vertices at coordinate axes with coordinates $(10,0,0)$, $(0,10,0)$ and $(0,0,10)$ (when we put any two coordinates to be zero in the equation we get that the third one has to be 10). The domain is the region of the space under this plane (including the points of the plane).