Find a symmetric point on a line I'm given a point $A (2; -6; 5)$, I need to find a symmetric point of A in regards of the straight line:
$$
2x+y+9z+5=0
$$
$$
x-3y+5z-4=0
$$
I began by making the line equation into parametric equation, which is:
$$
x=-32t-61
$$
$$
y=t
$$
$$
z=7t+13
$$
The problem I am facing, is that I don't know how should I move forward.
Edit: What I did now, is that I found a plane, on which the point A is and which's normal vector is the t values on the parametric equation. I found the intersection point of the line and the plane (by substituting the parametric equation's values into the plane equation) and then, because the point I needed to find, had to be symmetric to point A and the line, the point C (the symmetric point) was found with Cx=2Ox-Ax (where O is the intersection point of the plane and the line). With that logic, I found that the symmetric point of A in regards of the line is $C(2056/537; -5300/537; 2101/537)$, but the given answer is $C(4; 2; 3)$. Did I make a mistake or the given answer is wrong?
Plane equation which I got with $n * AP = 0$ is:
$$
-32x+y+7z+35=0
$$
By substituting the parametric values into it, I got:
$$
-32(-32t-61)+t+7(7t+13)+35=0
$$
which is
$$
1024t+1952+t+49t+91+35=0
$$
which is
$$
1074t = -2078
$$
where
$$
t = -1039/537 => y
$$
By checking the value of t, you can already see that the answer for the point wont be anywhere near as "beautiful" as the given answer.
 A: $\newcommand{\Vec}[1]{\mathbf{#1}}\DeclareMathOperator{\proj}{proj}$If $\ell$ is the line through some point $\Vec{x}_{0} = (x_{0}, y_{0}, z_{0})$ and having direction vector $\Vec{v} = (a, b, c)$, then a point $\Vec{x} = (x, y, z)$ decomposes as
$$
\Vec{x} = \Vec{x}_{0} + \underbrace{\Vec{x} - \Vec{x}_{0}}_{\Vec{w}}
= \Vec{x}_{0} + \underbrace{\proj_{\Vec{v}} \Vec{w}}_{\parallel \Vec{v}} + \underbrace{\Vec{w} - \proj_{\Vec{v}} \Vec{w}}_{\perp \Vec{v}},
$$
and the "symmetric point" is the image $R(\Vec{x})$ under reflection in $\ell$, namely
\begin{align*}
  R(\Vec{x})
  &= \Vec{x}_{0} + \proj_{\Vec{v}} \Vec{w} - (\Vec{w} - \proj_{\Vec{v}} \Vec{w}) \\
  &= \Vec{x}_{0} - \Vec{w} + 2\proj_{\Vec{v}} \Vec{w}.
\tag{1}
\end{align*}
This formula may be easier to use than your method once you have the line in parametric form.

Your parametric equations check out, i.e, the resulting line lies in each plane, so your answer looks plausible (unless the problem has been mis-copied): Take
$$
\Vec{x}_{0} = (-61, 0, 13),\qquad
\Vec{v} = (-32, 1, 7).
$$
Since $\Vec{x} = (2, -6, 5)$, you have $\Vec{w} = \Vec{x} - \Vec{x}_{0} = (63, -6, -8)$,
\begin{align*}
  \proj_{\Vec{v}} \Vec{w}
  &= \frac{\Vec{v} \cdot \Vec{w}}{\|\Vec{v}\|^{2}}\, \Vec{v}
  = \frac{-2016 - 6 - 56}{1024 + 1 + 49} (-32, 1, 7)
  = -\frac{1039}{537} (-32, 1, 7), \\
\Vec{w} - \proj_{\Vec{v}} \Vec{w}
  &= (63, -6, -8) + \frac{1039}{537} (-32, 1, 7),
\end{align*}
and
$$
R(\Vec{x}) = (-61, 0, 13) - (63, -6, -8) - \frac{2078}{537} (-32, 1, 7).
$$
