Answer
$$\eqalign{
& 3x + 2y + z = - 6 \cr
& {\text{Line: }}\frac{{x + 2}}{{ - 3}} = \frac{{y + 3}}{{ - 2}} = \frac{{z - 6}}{{ - 1}} \cr} $$
Work Step by Step
$$\eqalign{
& xy - z = 0,{\text{ }}\left( { - 2, - 3,6} \right) \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = xy - z \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = y,{\text{ }}{F_y}\left( {x,y,z} \right) = x,{\text{ }}{F_z}\left( {x,y,z} \right) = - 1,{\text{ }} \cr
& {\text{At the point }}\left( { - 2, - 3,6} \right) \cr
& {F_x}\left( { - 2, - 3,6} \right) = - 3,{\text{ }}{F_y}\left( { - 2, - 3,6} \right) = - 2,{\text{ }}{F_z}\left( { - 2, - 3,6} \right) = - 1,{\text{ }} \cr
& {\text{Direction numbers: }} - {\text{3}}, - 2, - 1 \cr
& {\text{An equation of the tangent plane at }}\left( { - 2, - 3,6} \right){\text{ is}} \cr
& - 3\left( {x + 2} \right) - 2\left( {y + 3} \right) - 1\left( {z - 6} \right) = 0 \cr
& - 3x - 6 - 2y - 6 - z + 6 = 0 \cr
& - 3x - 2y - z = 6 \cr
& 3x + 2y + z = - 6 \cr
& \cr
& {\text{The corresponding symmetric equations are:}} \cr
& {\text{Line: }}\frac{{x + 2}}{{ - 3}} = \frac{{y + 3}}{{ - 2}} = \frac{{z - 6}}{{ - 1}} \cr} $$
