Find the line symmetric to ell with respect to the plane π…

Find the line symmetric to $\ell$ with respect to the plane $\pi:\ x+y+4z-9=0$, where $\ell:\ \frac{x-1}{2}=\frac{y-2}{-4}=\frac{z+3}{5}$.

Explanation of Method

The mirror image of a line in a plane, using the piercing point and one reflected point.

Step 1: Piercing point

$\vec d=\{2,-4,5\}$ and $\vec n=\{1,1,4\}$, so $\vec d\cdot\vec n=18\neq0$ and $\ell$ meets $\pi$. With $x=1+2t,\ y=2-4t,\ z=-3+5t$: $18t-18=0\ \Longrightarrow\ t=1,\qquad P(3,-2,2)$.

Step 2: Reflect one point

For $Q(1,2,-3)$, substituting into the plane gives $1+2-12-9=-18$ and $|\vec n|^2=18$. The foot and the image are

\[ Q'=Q+\vec n=(2,3,1),\qquad Q^*=2Q'-Q=(3,4,5). \]

The direction is then $\overrightarrow{PQ^*}=\{0,6,3\}\parallel\{0,2,1\}$. The same comes from the reflection formula: $\vec d^*=\vec d-2\,\tfrac{\vec d\cdot\vec n}{|\vec n|^2}\vec n=\{0,-6,-3\}$.

Step 3: Result

\[\ell^*:\ \frac{x-3}{0}=\frac{y+2}{2}=\frac{z-2}{1},\quad\text{i.e. } x=3,\ \ y-2z+6=0\]

Checks: $Q^*$ gives $\tfrac{4+2}2=3=\tfrac{5-2}1$. For the angles with $\vec n$: \[\frac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}=\frac{18}{3\sqrt5\cdot3\sqrt2}=\frac{2}{\sqrt{10}},\qquad \frac{|\vec d^*\cdot\vec n|}{|\vec d^*|\,|\vec n|}=\frac{6}{\sqrt5\cdot3\sqrt2}=\frac{2}{\sqrt{10}} .\]

The two lines make equal angles with the plane. $\checkmark$

$P$ $Q$ $Q'$ $Q^*$ $\ell$ $\ell^*$

$\ell$ (red) pierces the plane at $P(3,-2,2)$. $Q(1,2,-3)$ is reflected through $Q'(2,3,1)$ to $Q^*(3,4,5)$, and $\ell^*=PQ^*$ (teal). Dashed parts lie below the plane.

Result

\[\boxed{\,\ell^*:\ \frac{x-3}{0}=\frac{y+2}{2}=\frac{z-2}{1},\quad\text{i.e. } x=3,\ \ y-2z+6=0\,}\]

Remark

Cleanup: the source's check reads "$3\neq3\ \checkmark$". Both ratios equal $3$.