Example 2. Find the coordinates of point $Q$, symmetric to point $P(-6,7,-9)$ with respect to the plane passing through points $A(1,3,-1), B(6,5,-2)$, and $C(0,-3,-5)$.
Solution. We will use the sketch of the problem (Fig. 4.13).

1) We will write the equation of the plane $\alpha=(A B C)$ passing through three points:

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-047.jpg?height=252&width=343&top_left_y=737&top_left_x=103)

Fig. 4.13 $(\alpha):\left|\begin{array}{ccc}x-1 & y-3 & z+1 \\ 5 & 2 & -1 \\ 1 & 6 & 4\end{array}\right|=0$.

We will skip the details since we performed a similar action in the previous problem. After expanding the determinant, we get the equation $(A B C): 2 x-3 y+4 z+11=0$.

2) We will write the equation of the line $l$ passing through point $P$ and perpendicular to $\alpha$. We take $\vec{l}=\vec{N}_{\alpha}=\{2,-3,4\}$;

$$
\frac{x+6}{2}=\frac{y-7}{-3}=\frac{z+9}{4} \text {, or } x=-6+2 t, y=7-3 t, z=-9+4 t
$$

3) We will determine the coordinates of point $O$ of intersection of $l$ and $\alpha$. We have: $2(-6+2 t)-3(7-3 t)+4(-9+4 t)+11=0,29 t=58, t=2$. After substituting $t=2$ into the parametric equations of the line, we get: $x=4-6=-2, \quad y=7-6=1, \quad z=-9+8=-1$.
4) Point $O(-2,1,-1)$ bisects the segment $P Q$, i.e., in particular, $x_{0}=\left(x_{P}+x_{Q}\right): 2$, or $x_{Q}=2 x_{0}-x_{P}$. We will use similar formulas for $y_{Q}$ and $z_{Q}$. We get $x_{Q}=-4+6=2, \quad y_{Q}=2-7=-5$, $z_{Q}=-2+9=7$.

Answer. $Q(2,-5,7)$