The d.r. of normal to the plane through $$\left( {1,\,0,\,0} \right),\,\left( {0,\,1,\,0} \right)$$     which makes an angle $$\frac{\pi }{4}$$ with plane $$x + y = 3$$   are :

Solution :
$$\eqalign{ & {\text{Equation of plane through}}\left( {1,\,0,\,0} \right){\text{ is}} \cr & a\left( {x - 1} \right) + by + cz = 0......\left( {\text{i}} \right) \cr & \left( {\text{i}} \right){\text{passes through}}\left( {0,\,1,\,0} \right) \cr & - a + b = 0 \cr & \Rightarrow b = a\,;\,{\text{Also,}} \cr & {\text{cos}}\,{45^ \circ } = \frac{{a + a}}{{\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \cr & \Rightarrow 2a = \sqrt {2{a^2} + {c^2}} \cr & \Rightarrow 2{a^2} = {c^2} \cr & \Rightarrow c = \sqrt 2 a \cr & {\text{So d}}{\text{.r of normal are }}a,\,a,\,\sqrt 2 a\,{\text{i}}{\text{.e}}{\text{., }}1,\,1,\,\sqrt 2 \cr} $$