Answer
$${\text{Surface area}} = 9\sqrt {14} \pi $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 12 + 2x - 3y \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {12 + 2x - 3y} \right] = 2 \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {12 + 2x - 3y} \right] = - 3 \cr
& {\text{The limits of the region }}R{\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \leqslant 9} \right\} \cr
& {\text{Changing the region to polar coordinates}} \cr
& R = \left\{ {\left( {r,\theta } \right):0 \leqslant r \leqslant 3,{\text{ 0}} \leqslant \theta \leqslant 2\pi } \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^{2\pi } {\int_0^3 {\sqrt {1 + {{\left[ 2 \right]}^2} + {{\left[ { - 3} \right]}^2}} rdrd\theta } } \cr
& = \int_0^{2\pi } {\int_0^3 {\sqrt {14} rdrd\theta } } \cr
& = \sqrt {14} \int_0^{2\pi } {\int_0^3 {rdrd\theta } } \cr
& {\text{Integrate with respect to }}r \cr
& = \sqrt {14} \int_0^{2\pi } {\left[ {\frac{{{r^2}}}{2}} \right]_0^3} d\theta \cr
& = \sqrt {14} \int_0^{2\pi } {\left( {\frac{9}{2}} \right)} d\theta \cr
& = \frac{9}{2}\sqrt {14} \int_0^{2\pi } {d\theta } \cr
& {\text{Integrate}} \cr
& {\text{Surface area}} = \frac{9}{2}\sqrt {14} \left( {2\pi } \right) \cr
& {\text{Surface area}} = 9\sqrt {14} \pi \cr} $$
