Origin point of a normal vector to a given point, from a given plane I have a plane $P: 2x + y - z -1 = 0.$
I have a point $A_1( 0, 1, 1).$
Question asks to find the coordinates of $A_2$, which is symetrical to $A_1$ with respect to $P$.
I tried using the formula for distance between a point and a plane(for $A_1$ and $P$) and with it, tried to find the coords of $A_2$, but that was unsuccesful.
I believe the last prospects are in finding a normal vector $N_1$, colinear to the normal to $P$ $n(2,1,-1)$, whose module is equal to the distance between $A_1$ and $P$, then multiplying by $-1$ and getting the answer.
I ask for help in finding this $N_1$ vector.
 A: The vector $v=(2,1,-1)$ is orthogonal to $P$. So, the point of $P$ which is closer to $A$ is the intersection of $P$ with the line passing through $A_1$ and with the direction given by $v$, which is the line$$\{(0,1,1)+t(2,1,-1)\mid t\in\Bbb R\}=\{(2t,t+1,-t+1)\mid t\in\Bbb R\}.$$And $(2t,t+1,-t+1)\in P$ if and only if $4t+(t+1)-(-t+1)-1=0$; in other words, if and only if $t=\frac16$. So, the intersection point is $I=\left(\frac13,\frac76,\frac56\right)$. And then\begin{align}A_2&=I-(A_1-I)\\&=2I-A_1\\&=\left(\frac23,\frac43,\frac23\right).\end{align}
A: Suppose (signed) distance of $A_1$ from plane $P$ is $D$. Then coordinates of its image $A_2$ will be given by
$$A_2 = A_1 -2D\hat{n}$$
Factor of $2$ is there since $A_2$ is at twice the distance from $A_1$ to the plane. Negative sign because $\vec{n}=(2,1,-1)$ points outward from the plane.  The unit vector $\hat{n}=\dfrac{1}{\sqrt{6}}(2,1,-1)$.
Signed distance $D$ of a point $(x_0,y_0,z_0)$ from a plane $ax+by+cz+d=0$ is given by
$$D=\frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}$$
Hence
$$A_2=(0,1,1)-2\cdot \Big(\frac{-1}{\sqrt{6}}\Big) \cdot \dfrac{1}{\sqrt{6}}(2,1,-1)=\left(\frac23,\frac43,\frac23\right)$$
A: You can take $n = (2, 1, -1)$ as your normal vector to the plane. Then, we can define $r(t) = A_1 + t\cdot n$. This is a line going through $A_1$ which goes in the direction of the normal vector to the plane.
If you can find the time $t_0$ where this line intersects the plane, continuing along the line for the same amount of time will allow you to find $A_2$. In other words, if $r(t_0) \in P$, then  $r(2t_0)$ will be $A_2$.
You can find $t_0$ by solving $n \cdot r(t) = 1$, which is just a linear equation.
