Reflect a plane by retaining its intersection and reflecting one point

With normal $(-1,1,2)$ and line direction $(5,2,-1)$, the absolute dot product is 5. Their acute angle is $\alpha=\cos^{-1}[5/(\sqrt{30}\sqrt6)]=68.1^\circ$(1d.p.). The line-plane angle is $90^\circ-\alpha=21.9^\circ$(1d.p.).

$\theta=\sin^{-1}[5/(\sqrt{30}\sqrt6)]=21.9^\circ$(1d.p.).

$BN:\mathbf r=(-1,0,-2)+\lambda(-1,1,2)$. At N,$(-1-\lambda,\lambda,-2+2\lambda)\cdot(-1,1,2)=3$. Thus $1+\lambda+\lambda-4+4\lambda=3$, λ=1, and $ON=(-2,1,0)$.

D=(0,1,1) lies on Π1 and $DB=(-1,-1,-3)$. Its normal component is $NB=\frac{DB\cdot(-1,1,2)}6(-1,1,2)=-(-1,1,2)$. Consequently $ON=OB-NB=(-1,0,-2)+(-1,1,2)=(-2,1,0)$ again.

$OB′=2ON-OB=2(-2,1,0)-(-1,0,-2)=(-3,2,2)$, hence B′=(−3,2,2).

$AB=(-2,2,-1)$ and $(5,2,-1)$ are parallel to Π2. Their cross product is $(0,-7,-14)=-7(0,1,2)$. Through B, Π2:$\mathbf r\cdot(0,1,2)=(-1,0,-2)\cdot(0,1,2)=-4$.

Π1:$-x+y+2z=3$ and Π2:$y+2z=-4$ give x=−7. Thus $l_2:\mathbf r=(-7,-4,0)+\mu(0,-2,1)$ and its Cartesian form is $x=-7$, $(y+4)/(-2)=z$.

Π3 contains l2, a fixed line of the mirror plane, and B′. With E=(−7,−4,0) on l2,$EB′=(-3,2,2)-(-7,-4,0)=(4,6,2)=2(2,3,1)$. Thus $\Pi_3:\mathbf r=(-3,2,2)+s(2,3,1)+t(0,-2,1)$, s,t real.