Answer
$$\theta \approx 77.4^\circ $$
Work Step by Step
$$\eqalign{
& {x^2} - {y^2} + z = 0,{\text{ }}\left( {1,2,3} \right) \cr
& {\text{Consider}} \cr
& F\left( {x,y,z} \right) = {x^2} - {y^2} + z \cr
& {\text{Calculating the partial derivatives}} \cr
& {F_x}\left( {x,y,z} \right) = 2x \cr
& {F_y}\left( {x,y,z} \right) = - 2y \cr
& {F_z}\left( {x,y,z} \right) = 1 \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr
& \nabla F\left( {x,y,z} \right) = {F_x}\left( {x,y,z} \right){\bf{i}} + {F_y}\left( {x,y,z} \right){\bf{j}} + {F_z}\left( {x,y,z} \right){\bf{k}} \cr
& \nabla F\left( {x,y,z} \right) = 2x{\bf{i}} - 2y{\bf{j}} + {\bf{k}} \cr
& {\text{Find the gradient }}\nabla F\left( {x,y,z} \right){\text{ at the point }}\left( {1,2,3} \right) \cr
& \nabla F\left( {1,2,3} \right) = 2\left( 1 \right){\bf{i}} - 2\left( 2 \right){\bf{j}} + {\bf{k}} \cr
& \nabla F\left( {1,2,3} \right) = 2{\bf{i}} - 4{\bf{j}} + {\bf{k}} \cr
& {\text{The angle of inclination of the tangent plane is}} \cr
& \cos \theta = \frac{{\left| {\nabla F\left( {1,2,3} \right) \cdot {\bf{k}}} \right|}}{{\left\| {\nabla F\left( {1,2,3} \right)} \right\|}} \cr
& \cos \theta = \frac{{\left| {\left( {2{\bf{i}} - 4{\bf{j}} + {\bf{k}}} \right) \cdot {\bf{k}}} \right|}}{{\left\| {2{\bf{i}} - 4{\bf{j}} + {\bf{k}}} \right\|}} \cr
& \cos \theta = \frac{{\left| 1 \right|}}{{\sqrt {4 + 16 + 1} }} \cr
& \cos \theta = \frac{1}{{\sqrt {21} }} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {21} }}} \right) \cr
& \theta \approx 77.4^\circ \cr} $$
