<s>[INST] Symmetric point of a point wrt a line I am trying to solve the following question.

Let $d$ be the
$$
\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{5},
$$
and let $P$ be the point of coordinates $(4,3,10)$. Find the coordinates of the symmetric point $P^{\prime}$ of $P$ with respect to $d$.

Book has the following solution.
Let $M\left(x_0, y_0, z_0\right)$ be the midpoint of $P P^{\prime}$. Then $M \in d, P M \perp$ d, i.e.,
$$
\begin{gathered}
\frac{x_0-1}{2}=\frac{y_0-2}{4}=\frac{z_0-3}{5} \\
2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0 .
\end{gathered}
$$
This yields $x_0=129 / 41, y_0=258 / 41, z_0=343 / 41$. Now, the coordinates of $P^{\prime}$ are given by $\left(P, M ; P^{\prime}\right)=-2$ (which is the simple ratio), and we have $P^{\prime}(-94 / 41$, $-393 / 41,-276 / 41)$
However, I don't understand why he takes $2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0$ instead of $2\left(x_0-4\right)+4\left(y_0-3\right)+5\left(z_0-10\right)=0$. Since $PM$ is orthogonal to $d$, I am in $PM\cdot (2,4,5)$ must be equal to zero since $(2,4,5)$ is direction vector of the line. If we use the dot product that I wrote, I calculated $x_0=135/13, y_0=-270/13$.
Is my solution wrong?
I appreciate any help in advance.
 [/INST] The equations are
$ (x_0, y_0, z_0) = (1,2,3) + t (2, 4, 5) $
$PM = M - P = (1, 2, 3) - (4, 3, 10) + t (2, 4, 5)  = (-3, -1, -7) + t (2, 4, 5)$
$ PM \perp d $ , so
$ \bigg( (-3, -1, -7) + t(2, 4, 5) \bigg) \cdot (2, 4, 5) = 0 $
So that
$ -45 + 45 t = 0 $
$\Longrightarrow t = 1 $
Thus $M = (1,2,3) + (2,4,5) = (3, 6, 8) $
$M = \dfrac{1}{2} (P + P') $
So
$P' = 2 M - P = 2 (3, 6, 8) - (4,3,10) = (2, 9, 6) $
</s>