Answer
$$12x - 6y - 4z = 22$$
Work Step by Step
$$\eqalign{
& x{y^2} + 3x - {z^2} = 8,{\text{ }}\left( {1, - 3,2} \right){\text{ }} \cr
& x{y^2} + 3x - {z^2} - 8 = 0{\text{ }} \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = x{y^2} + 3x - {z^2} - 8 \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {x{y^2} + 3x - {z^2} - 8} \right] = {y^2} + 3 \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {x{y^2} + 3x - {z^2} - 8} \right] = 2xy \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {x{y^2} + 3x - {z^2} - 8} \right] = - 2z \cr
& {\text{At the point }}\left( {1, - 3,2} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {1, - 3,2} \right) = {\left( { - 3} \right)^2} + 3 = 12 \cr
& {F_y}\left( {1, - 3,2} \right) = 2\left( 1 \right)\left( { - 3} \right) = - 6 \cr
& {F_z}\left( {1, - 3,2} \right) = - 2\left( 2 \right) = - 4 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& 12\left( {x - 1} \right) - 6\left( {y + 3} \right) - 4\left( {z - 2} \right) = 0 \cr
& {\text{Simplifying}} \cr
& 12x - 12 - 6y - 18 - 4z + 8 = 0 \cr
& 12x - 6y - 4z - 22 = 0 \cr
& 12x - 6y - 4z = 22 \cr} $$
