Answer
$Area\left( S \right) = \frac{5}{4}\sqrt {29} $
Work Step by Step
From the equation of the plane $2x + 3y + 4z = 28$, we can write $z = \frac{1}{4}\left( {28 - 2x - 3y} \right)$.
So, the plane can be parametrized by $G\left( {x,y} \right) = \left( {x,y,\frac{1}{4}\left( {28 - 2x - 3y} \right)} \right)$.
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0, - \frac{1}{2}} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1, - \frac{3}{4}} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - \frac{1}{2}}\\
0&1&{ - \frac{3}{4}}
\end{array}} \right| = \frac{1}{2}{\bf{i}} + \frac{3}{4}{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{3}{4}} \right)}^2} + 1} = \frac{{\sqrt {29} }}{4}$
By Theorem 1, the surface area of $S$ is
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$Area\left( S \right) = \frac{{\sqrt {29} }}{4}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$
Since $Area\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y = 5$, so the area of the portion of the plane is
$Area\left( S \right) = \frac{5}{4}\sqrt {29} $
