Answer
$$\theta \approx 86.03^\circ $$
Work Step by Step
$$\eqalign{
& 3{x^2} + 2{y^2} - z = 15,{\text{ }}\left( {2,2,5} \right) \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = 3{x^2} + 2{y^2} - z - 15 \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 6x{\text{ }} \cr
& {F_y}\left( {x,y,z} \right) = 4y{\text{ }} \cr
& {F_z}\left( {x,y,z} \right) = - 1{\text{ }} \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr
& \nabla F\left( {x,y,z} \right) = {F_x}\left( {x,y,z} \right){\bf{i}} + {F_y}\left( {x,y,z} \right){\bf{j}} + {F_z}\left( {x,y,z} \right){\bf{k}} \cr
& \nabla F\left( {x,y,z} \right) = 6x{\bf{i}} + 4y{\bf{j}} - {\bf{k}} \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right){\text{ at the point }}\left( {2,2,5} \right) \cr
& \nabla F\left( {2,2,5} \right) = 6\left( 2 \right){\bf{i}} + 4\left( 2 \right){\bf{j}} - {\bf{k}} \cr
& \nabla F\left( {2,2,5} \right) = 12{\bf{i}} + 8{\bf{j}} - {\bf{k}} \cr
& {\text{The angle of inclination of the tangent plane is}} \cr
& \cos \theta = \frac{{\left| {\nabla F\left( {2,2,5} \right) \cdot {\bf{k}}} \right|}}{{\left\| {\nabla F\left( {2,2,5} \right)} \right\|}} \cr
& \cos \theta = \frac{{\left| {\left( {12{\bf{i}} + 8{\bf{j}} - {\bf{k}}} \right) \cdot {\bf{k}}} \right|}}{{\left\| {12{\bf{i}} + 8{\bf{j}} - {\bf{k}}} \right\|}} \cr
& \cos \theta = \frac{{\left| {0 + 0 - 1} \right|}}{{\sqrt {144 + 64 + 1} }} \cr
& \cos \theta = \frac{1}{{\sqrt {209} }} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {209} }}} \right) \cr
& \theta \approx 86.03^\circ \cr} $$
