Answer
$$48\sqrt {14} $$
Work Step by Step
$$\eqalign{
& z = 24 - 3x - 2y \cr
& f\left( {x,y} \right) = 24 - 3x - 2y \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {24 - 3x - 2y} \right] = - 3 \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {24 - 3x - 2y} \right] = - 2 \cr
& {\text{Let }}z = 0 \cr
& 0 = 24 - 3x - 2y \cr
& 2y = - 3x + 24 \cr
& y = - \frac{3}{2}x + 12 \cr
& {\text{The limits of the region }}R{\text{ }}\left( {{\text{shown below}}} \right){\text{ are:}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant - \frac{3}{2}x + 12,{\text{ 0}} \leqslant x \leqslant 8} \right\} \cr
& \cr
& {\text{The area of the surface }}S{\text{ is given by}} \cr
& {\text{Surface area}} = \iint\limits_R {\sqrt {1 + {{\left[ {{f_x}\left( {x,y} \right)} \right]}^2} + {{\left[ {{f_y}\left( {x,y} \right)} \right]}^2}} dA} \cr
& = \int_0^8 {\int_0^{ - \frac{3}{2}x + 12} {\sqrt {1 + {{\left[ { - 3} \right]}^2} + {{\left[ { - 2} \right]}^2}} dydx} } \cr
& = \int_0^8 {\int_0^{ - \frac{3}{2}x + 12} {\sqrt {14} dydx} } \cr
& {\text{Integrate with respect to }}y \cr
& = \sqrt {14} \int_0^8 {\left( { - \frac{3}{2}x + 12} \right)} dx \cr
& = \sqrt {14} \left[ { - \frac{3}{4}{x^2} + 12x} \right]_0^8 \cr
& = \sqrt {14} \left[ { - \frac{3}{4}{{\left( 8 \right)}^2} + 12\left( 8 \right)} \right] - \sqrt {14} \left[ { - \frac{3}{4}{{\left( 0 \right)}^2} + 12\left( 0 \right)} \right] \cr
& = 48\sqrt {14} \cr} $$
