Answer
$$\eqalign{
& 2x + 4y + z = 14 \cr
& {\text{Line: }}\frac{{x - 1}}{2} = \frac{{y - 2}}{4} = \frac{{z - 4}}{1} \cr} $$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} + z = 9,{\text{ }}\left( {1,2,4} \right) \cr
& {x^2} + {y^2} + z - 9 = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = {x^2} + {y^2} + z - 9 \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 2x,{\text{ }}{F_y}\left( {x,y,z} \right) = 2y,{\text{ }}{F_z}\left( {x,y,z} \right) = 1,{\text{ }} \cr
& {\text{At the point }}\left( {1,2,4} \right) \cr
& {F_x}\left( {1,2,2} \right) = 2,{\text{ }}{F_y}\left( {1,2,2} \right) = 4,{\text{ }}{F_z}\left( {1,2,2} \right) = 1,{\text{ }} \cr
& {\text{Direction numbers: }}2,4,1 \cr
& {\text{An equation of the tangent plane at }}\left( {1,2,4} \right){\text{ is}} \cr
& {F_x}\left( {1,2,4} \right)\left( {x - 1} \right) + {F_y}\left( {1,2,4} \right)\left( {y - 2} \right) + {F_z}\left( {1,2,4} \right)\left( {z - 2} \right) = 0 \cr
& 2\left( {x - 1} \right) + 4\left( {y - 2} \right) + \left( {z - 4} \right) = 0 \cr
& 2x - 2 + 4y - 8 + z - 4 = 0 \cr
& 2x + 4y + z = 14 \cr
& {\text{The corresponding symmetric equations are:}} \cr
& {\text{Line: }}\frac{{x - 1}}{2} = \frac{{y - 2}}{4} = \frac{{z - 4}}{1} \cr} $$
