Answer
$$3x + 4y - 5z = 0$$
Work Step by Step
$$\eqalign{
& z = \sqrt {{x^2} + {y^2}} ,{\text{ }}\left( {3,4,5} \right) \cr
& z - \sqrt {{x^2} + {y^2}} = 0 \cr
& {\text{Considering }} \cr
& F\left( {x,y,z} \right) = z - \sqrt {{x^2} + {y^2}} \cr
& {\text{Calculate the partial derivatives }} \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - \sqrt {{x^2} + {y^2}} } \right] = - \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - \sqrt {{x^2} + {y^2}} } \right] = - \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - \sqrt {{x^2} + {y^2}} } \right] = 1 \cr
& {\text{At the point }}\left( {3,4,5} \right){\text{ the partial derivatives are}} \cr
& {F_x}\left( {3,4,5} \right) = - \frac{3}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }} = - \frac{3}{5} \cr
& {F_y}\left( {3,4,5} \right) = - \frac{4}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }} = - \frac{4}{5} \cr
& {F_z}\left( {3,4,5} \right) = 1 \cr
& {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr
& {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr
& + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr
& {\text{Substituting}} \cr
& - \frac{3}{5}\left( {x - 3} \right) - \frac{4}{5}\left( {y - 4} \right) + \left( {z - 5} \right) = 0 \cr
& {\text{Simplifying}} \cr
& - \frac{3}{5}x + \frac{9}{5} - \frac{4}{5}y + \frac{{16}}{5} + z - 5 = 0 \cr
& - \frac{3}{5}x - \frac{4}{5}y + z = 0 \cr
& 3x + 4y - 5z = 0 \cr} $$
