Answer
$$x - 4y + 2z = 18$$
Work Step by Step
$$\eqalign{
& {x^2} + 4{y^2} + {z^2} = 36,{\text{ }}\left( {2, - 2,4} \right) \cr
& {x^2} + 4{y^2} + {z^2} - 36 = 0 \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = {x^2} + 4{y^2} + {z^2} - 36 \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + 4{y^2} + {z^2} - 36} \right] = 2x \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 4{y^2} + {z^2} - 36} \right] = 8y \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^2} + 4{y^2} + {z^2} - 36} \right] = 2z \cr
& {\text{At the point }}\left( {2, - 2,4} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {2, - 2,4} \right) = 2\left( 2 \right) = 4 \cr
& {F_y}\left( {2, - 2,4} \right) = 8\left( { - 2} \right) = - 16 \cr
& {F_z}\left( {2, - 2,4} \right) = 2\left( 4 \right) = 8 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& 4\left( {x - 2} \right) - 16\left( {y + 2} \right) + 8\left( {z - 4} \right) = 0 \cr
& {\text{Simplifying}} \cr
& 4x - 8 - 16y - 32 + 8z - 32 = 0 \cr
& 4x - 16y + 8z - 72 = 0 \cr
& 4x - 16y + 8z = 72 \cr
& x - 4y + 2z = 18 \cr} $$