Answer
$$z = 4$$
Work Step by Step
$$\eqalign{
& z = - 9 + 4x - 6y - {x^2} - {y^2},{\text{ }}\left( {2, - 3,4} \right) \cr
& z + 9 - 4x + 6y + {x^2} + {y^2} = 0 \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = z + 9 - 4x + 6y + {x^2} + {y^2} \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z + 9 - 4x + 6y + {x^2} + {y^2}} \right] = - 4 + 2x \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z + 9 - 4x + 6y + {x^2} + {y^2}} \right] = 6 + 2y \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z + 9 - 4x + 6y + {x^2} + {y^2}} \right] = 1 \cr
& {\text{At the point }}\left( {2, - 3,4} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {2, - 3,4} \right) = - 4 + 2\left( 2 \right) = 0 \cr
& {F_y}\left( {2, - 3,4} \right) = 6 + 2\left( { - 3} \right) = 0 \cr
& {F_z}\left( {2, - 3,4} \right) = 1 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& {F_x}\left( {2, - 3,4} \right)\left( {x - 2} \right) + {F_y}\left( {2, - 3,4} \right)\left( {y + 3} \right) \cr
& + {F_z}\left( {2, - 3,4} \right)\left( {z - 4} \right) = 0 \cr
& 0\left( {x - 2} \right) + 0\left( {y + 3} \right) + \left( {z - 4} \right) = 0 \cr
& {\text{Simplifying}} \cr
& z - 4 = 0 \cr
& z = 4 \cr} $$
