Planes are drawn so as to make an angle $60 ^\circ$ with the line $x = y = z$ and the angle of $45^\circ$ with the line $x=0=y-z$. Show that all these planes make an angle $60^\circ$ with the plane $x=0$.
I am assuming the required plane be of the form $ax+by+cz=d$ and Looking for hints to proceed further. any help is appreciated.
Let $\vec{n}(a,b,c)$ be a vector normal of the plane.
Thus, $$\frac{\sqrt3}{2}=\frac{|a+b+c|}{\sqrt{a^2+b^2+c^2}\sqrt3}$$ and $$\frac{1}{\sqrt2}=\frac{|b+c|}{\sqrt{a^2+b^2+c^2}\sqrt2}$$ and we need to prove that $$\frac{1}{2}=\frac{|a|}{\sqrt{a^2+b^2+c^2}}.$$ We obtain: $$3|b+c|=2|a+b+c|,$$ which gives $$3(b+c)=2(a+b+c)$$ or $$b+c=2a$$ or $$a+b+c=3a,$$ which gives $$|a+b+c|=3|a|$$ and indeed, $$\frac{1}{2}=\frac{|a|}{\sqrt{a^2+b^2+c^2}}.$$ Also, we have $$3(b+c)=-2(a+b+c)$$ or $$5(a+b+c)=3a$$ or $$5|a+b+c|=3|a|,$$ which gives $$\frac{5}{2}=\frac{|a|}{\sqrt{a^2+b^2+c^2}},$$ which is impossible.
Done!