Answer
a) 24
b) See below for explanation.
Work Step by Step
$$
g(x,y,z)=x^{3}y^{2}z\sqrt {10-x-y-z}
$$
(a)
$$
\begin{split}
g(1,2,3) &=(1)^{3} .(2)^{2}. (3)\sqrt {10-(1)-(2)-(3)} \\
&=12\sqrt {4}\\
&=24
\end{split}
$$
(b)
The expression for $f$ makes sense if the quantity under the square root sign is non-negative. So the domain of $f$ is
$$
\begin{split}
D &=\left\{ (x,y,z ) : 10-x-y-z \geq 0 \right\} \\
& =\left\{ (x,y,z ) : z \leq 10-x-y \right\}
\end{split}
$$
The inequality $z \leq 10-x-y $, describes the points on or below the plane $ z +x+y= 10 $.
