Transform a plane after finding intersections and parallel offsets

The line direction is $(7,1,-2)$ and plane normal $(5,3,b)$. A unique intersection requires $35+3-2b\ne0$, hence $b\ne19$.

$l:\mathbf r=(9,-3,-1)+\lambda(7,1,-2)$ and $p:\mathbf r\cdot(5,3,1)=-1$. Substitution gives $35+36\lambda=-1$, λ=−1, hence N=(2,−4,1).

$AN=(2,-4,1)-(-1,-11,2)=(3,7,-1)$. Thus distance=$|AN\cdot(5,3,1)|/\sqrt{25+9+1}=35/\sqrt{35}=\sqrt{35}$.

Let $p_1:\mathbf r\cdot(5,3,1)=k$. Then $2\sqrt{35}=|k-(-1)|/\sqrt{35}$, so $|k+1|=70$ and k=69 or −71. Either equation is a possible answer.

Choose B with A the midpoint of NB:$OB=2OA-ON=2(-1,-11,2)-(2,-4,1)=(-4,-18,3)$. The parallel plane through B has constant $(-4,-18,3)\cdot(5,3,1)=-20-54+3=-71$. This uses a point reflection of the whole plane through A; N is not asserted to be A’s perpendicular foot.

The combined image coordinates are $X=x-3$, $Y=y/2$, $Z=-z$. Thus the original coordinates are $x=X+3$, $y=2Y$, $z=-Z$. Substituting in $5x+3y+z+1=0$ gives $5(X+3)+6Y-Z+1=0$. Dropping capital labels, p2:$5x+6y-z=-16$.