Example 3. Find the coordinates of point $Q$, symmetric to point $P(1,3,2)$ with respect to the line $A B$, where $A(1,2,-6), B(7,-7,6)$.
Solution. 1) We have $\overrightarrow{A B}=\{6,-9,12\}$. We take $\vec{l}=\{2,-3,4\}$;

$$
(AB): \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+6}{4}
$$

or

$$
x=1+2 t, \quad y=2-3 t, \quad z=-6+4 t
$$

2) The equation of the plane $\alpha$, passing through $P$ and perpendicular to $AB$, is (Fig. 4.14)

$(\alpha): 2(x-1)-3(y-3)+4(z-2)=0$.

3) We find the coordinates of the point $O$ of intersection of $AB$ and $\alpha: 2(1+2 t-1)-3(2-$ $-3 t-3)+4(4 t-6-2)=0, t=1 ; x=3$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-047.jpg?height=281&width=395&top_left_y=183&top_left_x=2090)

Fig. 4.14 $y=-1, z=-2$. Thus, $O(3,-1,-2)$.

4. The coordinates of $Q$ are calculated using the previously used formulas: $x_{Q}=2 x_{0}-x_{P}$. We get $x_{Q}=5, y_{Q}=-5, z_{Q}=-6$.

Answer. $Q(5,-5,-6)$.