#### Answer

$h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$;
set of points: {$(x,y) | 2x+3y\geq 6$}

#### Work Step by Step

As per the given values; $h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$
The term $(2x+3y-6)^2$ is a polynomial and will be always continuous and the term $ \sqrt{ 2x+3y-6}$ is a squared root which only defines for the non-negative values.
Thus, $(2x+3y-6)\geq 0$
or, $(2x+3y)\geq 6$
or, $3y\geq -2x+6$
or, $ y\geq \dfrac{-2}{3}x+2$
Hence,
$h(x,y)=(2x+3y-6)^2+\sqrt{ 2x+3y-6}$;
set of points: {$(x,y) | 2x+3y\geq 6$}