Reflecting A Point Through A Plane

First, we construct the line passing through $(-6, 2, 3)$ that is perpendicular to the given plane. Since the normal vector to the plane is $\langle3, -4, 5\rangle$ , we have that this line is

$\begin{aligned} x &= 3t - 6 \\ y &= -4t + 2 \\ z &= 5t + 3 \end{aligned}$

Next, we'll plug this equation into the equation of the plane to find the point of intersection:

$3(3t - 6) - 4(-4t + 2) + 5(5t + 3) - 9 = 0 \longrightarrow t = \frac{2}{5}$

Plugging this value of $t$ back into the equation of the line yields the point $\left(-\dfrac{24}{5}, \dfrac{2}{5}, 5\right)$ . The vector from this point to $P$ is $\langle -\dfrac{6}{5}, \dfrac{8}{5}, -2\rangle$ . So we negate this vector and add it to the point $\left(-\dfrac{24}{5}, \dfrac{2}{5}, 5\right)$ to get the reflected point.

$\langle \dfrac{6}{5}, -\dfrac{8}{5}, 2\rangle + \left(-\dfrac{24}{5}, \dfrac{2}{5}, 5\right) = \left(-\frac{18}{5}, -\frac{6}{5}, 7\right)$

Adding the $x$ -, $y$ -, and $z$ -coordinates gives $\frac{11}{5} = 2.2$

Alternative method that doesn't intersect a line with the plane.

$\delta(P,\text{plane})=\dfrac{\big|3·(-6)-4·2+5·3-9\big|}{\sqrt{3^2+(-4)^2+5^2}}=2\sqrt2$ . We want to move twice this distance along the vector that is perpendicular to the plane $\vec{n}=\begin{pmatrix}3\\-4\\5\end{pmatrix}$ , that is $\vec{PQ}=\pm\dfrac{2·2\sqrt2}{5\sqrt2}\begin{pmatrix}3\\-4\\5\end{pmatrix}=\pm\begin{pmatrix}\frac{12}5\\\frac{-16}5\\4\end{pmatrix}$ .

That give two possibilities for $Q$ : $Q_1(\frac{-18}5,\frac{-6}5,7)$ and $Q_2(\frac{-42}5,\frac{26}5,-1)$ . To find the correct one, we find the middles $M_1$ and $M_2$ of $PQ_1$ and $PQ_2$ and see which one is on the plane.

$M_1(\frac{-24}5,\frac25,5)$ is on the plane, so the answer is $Q=Q_1$ . ( $M_2(\frac{-36}5,\frac{18}5,1)$ is not on the plane.)