Answer
$$\eqalign{
& 10x + 5y + 2z = 30 \cr
& {\text{Line: }}\frac{{x - 1}}{{10}} = \frac{{y - 2}}{5} = \frac{{z - 5}}{2} \cr} $$
Work Step by Step
$$\eqalign{
& xyz = 10,{\text{ }}\left( {1,2,5} \right) \cr
& xyz - 10 = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = xyz - 10 \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = yz,{\text{ }}{F_y}\left( {x,y,z} \right) = xz,{\text{ }}{F_z}\left( {x,y,z} \right) = xy,{\text{ }} \cr
& {\text{At the point }}\left( {1,2,5} \right) \cr
& {F_x}\left( {1,2,5} \right) = 10,{\text{ }}{F_y}\left( {1,2,5} \right) = 5,{\text{ }}{F_z}\left( {1,2,5} \right) = 2,{\text{ }} \cr
& {\text{Direction numbers: 10}},5,2 \cr
& {\text{An equation of the tangent plane at }}\left( {1,2,5} \right){\text{ is}} \cr
& {F_x}\left( {1,2,5} \right)\left( {x - 3} \right) + {F_y}\left( {1,2,5} \right)\left( {y - 2} \right) + {F_z}\left( {1,2,5} \right)\left( {z - 5} \right) = 0 \cr
& 10\left( {x - 1} \right) + 5\left( {y - 2} \right) + 2\left( {z - 5} \right) = 0 \cr
& 10x - 10 + 5y - 10 + 2z - 10 = 0 \cr
& 10x + 5y + 2z = 30 \cr
& \cr
& {\text{The corresponding symmetric equations are:}} \cr
& {\text{Line: }}\frac{{x - 1}}{{10}} = \frac{{y - 2}}{5} = \frac{{z - 5}}{2} \cr} $$
