Write the equation of the plane through the intersection of…

Write the equation of the plane through the intersection of $x-y+z+1=0$ and $x+y-z-1=0$ making an angle of $60^\circ$ with the first plane.

Step 1: A simpler pencil

Add and subtract the equations: $2x=0,\qquad -2y+2z+2=0$.

So the common line is $\ell:\ x=0,\ y-z-1=0$, and the pencil can be written as

\[ \alpha x+\beta(y-z-1)=0,\qquad \vec n=\{\alpha,\beta,-\beta\}. \]

Step 2: Angle condition

With $\vec n_1=\{1,-1,1\}$: \[\frac{|\alpha-2\beta|}{\sqrt3\,\sqrt{\alpha^2+2\beta^2}}=\frac12\ \Longrightarrow\ 4(\alpha-2\beta)^2=3(\alpha^2+2\beta^2) .\]

\[ \alpha^2-16\alpha\beta+10\beta^2=0\ \Longrightarrow\ \frac{\alpha}{\beta}=8\pm\sqrt{54}=8\pm3\sqrt6 . \]

The value $\beta=0$, i.e. the plane $x=0$, is not a solution.

\[(8+3\sqrt6)\,x+y-z-1=0\qquad\text{and}\qquad(8-3\sqrt6)\,x+y-z-1=0\]

Step 3: Comparison with the source

The source uses $(x-y+z+1)+\lambda(x+y-z-1)=0$ and gets $5\lambda^2+18\lambda-27=0$, so $\lambda=\tfrac{-9\pm6\sqrt6}{5}$. Its planes

\[ (-4\pm6\sqrt6)x+(-14\pm6\sqrt6)y+(14\mp6\sqrt6)z+(14\mp6\sqrt6)=0 \] are the same two planes. For example, dividing by $-14\pm6\sqrt6$ gives $\tfrac{-4\pm6\sqrt6}{-14\pm6\sqrt6}=8\pm3\sqrt6$ for the $x$-coefficient. The source's result is correct; the new form is much simpler.

Numerical check: both planes give $\varphi=60.000^\circ$.