Answer
$$\eqalign{
& 6x - 4y - z = 5 \cr
& {\text{Line: }}\frac{{x - 3}}{6} = \frac{{y - 2}}{{ - 4}} = \frac{{z - 5}}{{ - 1}} \cr} $$
Work Step by Step
$$\eqalign{
& z = {x^2} - {y^2},{\text{ }}\left( {3,2,5} \right) \cr
& {x^2} - {y^2} - z = 0 \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = {x^2} - {y^2} - z \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 2x,{\text{ }}{F_y}\left( {x,y,z} \right) = - 2y,{\text{ }}{F_z}\left( {x,y,z} \right) = - 1,{\text{ }} \cr
& {\text{At the point }}\left( {3,2,5} \right) \cr
& {F_x}\left( {3,2,5} \right) = 6,{\text{ }}{F_y}\left( {3,2,5} \right) = - 4,{\text{ }}{F_z}\left( {3,2,5} \right) = - 1,{\text{ }} \cr
& {\text{Direction numbers: 6}}, - 4, - 1 \cr
& {\text{An equation of the tangent plane at }}\left( {3,2,5} \right){\text{ is}} \cr
& {F_x}\left( {3,2,5} \right)\left( {x - 3} \right) + {F_y}\left( {3,2,5} \right)\left( {y - 2} \right) + {F_z}\left( {3,2,5} \right)\left( {z - 5} \right) = 0 \cr
& 6\left( {x - 3} \right) - 4\left( {y - 2} \right) - 1\left( {z - 5} \right) = 0 \cr
& 6x - 18 - 4y + 8 - z + 5 = 0 \cr
& 6x - 4y - z = 5 \cr
& \cr
& {\text{The corresponding symmetric equations are:}} \cr
& {\text{Line: }}\frac{{x - 3}}{6} = \frac{{y - 2}}{{ - 4}} = \frac{{z - 5}}{{ - 1}} \cr} $$
