Answer
${{\bf{T}}_u} = \left( {2,1,3} \right)$
${{\bf{T}}_v} = \left( {1, - 4,0} \right)$
${\bf{N}}\left( {1,4} \right) = 12{\bf{i}} + 3{\bf{j}} - 9{\bf{k}}$
The equation of the tangent plane:
$4x + y - 3z = 0$
Work Step by Step
We have $G\left( {u,v} \right) = \left( {2u + v,u - 4v,3u} \right)$. So,
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2,1,3} \right)$
${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {1, - 4,0} \right)$
${\bf{N}}\left( {1,4} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&1&3\\
1&{ - 4}&0
\end{array}} \right| = 12{\bf{i}} + 3{\bf{j}} - 9{\bf{k}}$
The equation of the tangent plane to the surface at the point $G\left( {1,4} \right) = \left( {6, - 15,3} \right)$ is
$\left( {x - 6,y + 15,z - 3} \right)\cdot\left( {12,3, - 9} \right) = 0$
$12\left( {x - 6} \right) + 3\left( {y + 15} \right) - 9\left( {z - 3} \right) = 0$
$4\left( {x - 6} \right) + y + 15 - 3\left( {z - 3} \right) = 0$
The equation can be written as
$4x + y - 3z = 0$