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