Find the plane through the line x+5y+z=0, x-z+4=0 making an…
Find the plane through the line $x+5y+z=0,\ x-z+4=0$ making an angle of $45^\circ$ with $x-4y-8z+12=0$.
Explanation of Method
Planes through a line that make an angle of 45° with a given plane.
Step 1: Pencil
Use $\alpha(x+5y+z)+\beta(x-z+4)=0$, with normal $\{\alpha+\beta,\ 5\alpha,\ \alpha-\beta\}$. The given normal is $\{1,-4,-8\}$, of length $9$: \[\frac{|9\beta-27\alpha|}{9\sqrt{27\alpha^2+2\beta^2}}=\frac1{\sqrt2}\ \Longrightarrow\ 2(\beta-3\alpha)^2=27\alpha^2+2\beta^2\ \Longrightarrow\ \alpha(4\beta+3\alpha)=0 .\]
Step 2: Two solutions
$\beta=-\tfrac34\alpha$ (i.e. $\lambda=-\tfrac34$ in the source) gives
\[x+20y+7z-12=0\]
$\alpha=0$ gives the member $\lambda=\infty$, the second plane itself: $x-z+4=0$.
Check: for $\{1,0,-1\}\cdot\{1,-4,-8\}=9$, $\cos\varphi=\tfrac{9}{\sqrt2\cdot9}=\tfrac1{\sqrt2}$. $\checkmark$
Result
\[\boxed{\,x+20y+7z-12=0\,}\]
\[\boxed{\,x-z+4=0\,}\]
Remark
Correction: the source writes the pencil as $(x+5y+z)+\lambda(x-z+4)=0$, which misses the member $x-z+4=0$ ($\lambda\to\infty$). This member also makes $45^\circ$ with the given plane, so there are two solutions. The squared equation became linear precisely because one root went to infinity.