Answer
$$\eqalign{
& x + y + z = 9 \cr
& {\text{Line: }}x - 3 = y - 3 = z - 3 \cr} $$
Work Step by Step
$$\eqalign{
& x + y + z = 9,{\text{ }}\left( {3,3,3} \right) \cr
& x + y + z - 9 = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = x + y + z - 9 \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 1,{\text{ }}{F_y}\left( {x,y,z} \right) = 1,{\text{ }}{F_z}\left( {x,y,z} \right) = 1,{\text{ }} \cr
& {\text{At the point }}\left( {3,3,3} \right) \cr
& {F_x}\left( {3,3,3} \right) = 1,{\text{ }}{F_y}\left( {3,3,3} \right) = 1,{\text{ }}{F_z}\left( {3,3,3} \right) = 1,{\text{ }} \cr
& {\text{Direction numbers: }}1,1,1 \cr
& {\text{An equation of the tangent plane at }}\left( {3,3,3} \right){\text{ is}} \cr
& {F_x}\left( {3,3,3} \right)\left( {x - 3} \right) + {F_y}\left( {3,3,3} \right)\left( {y - 3} \right) + {F_z}\left( {3,3,3} \right)\left( {z - 3} \right) = 0 \cr
& \left( {x - 3} \right) + \left( {y - 3} \right) + \left( {z - 3} \right) = 0 \cr
& x - 3 + y - 3 + z - 3 = 0 \cr
& x + y + z = 9 \cr
& {\text{The corresponding symmetric equations are:}} \cr
& {\text{Line: }}\frac{{x - 3}}{1} = \frac{{y - 3}}{1} = \frac{{z - 3}}{1} \cr
& {\text{Line: }}x - 3 = y - 3 = z - 3 \cr} $$
