Find the symmetrical point of $(1,2,3)$ and orthogonal to the plane $x+y+z=3$.
I know that $\vec{n} = {1,1,1}$ is a normal vector for the plane $x+y+z = 3$. Then the vector form of the line going through $(1,2,3)$ would be $\vec{r} = (1, 2, 3) + t(1,1,1)$. Two point is symmetric if their distance from the origin is the same, so we must have that $d = \sqrt{1 +4 + 9} = \sqrt{14} = \sqrt{(1-t)^2 + (2+t) ^2 + (3 + t)^2}$.
$\implies 3t^2 + 8t = 0$
$\implies t = \frac{-8 \pm \sqrt{64}}{6} \implies t = -\frac{8}{3}$
So the point that is symmetric to $(1,2,3)$ and orthogonal to the plane $x+y+x=3$ is $(\frac{-5}{3}, \frac{-1}{3}, \frac{1}{3})$
Is the above solution I have given correct?
$\sqrt {1^2 + 2^2 + 3^2}$ is the distance your point is from the origin not from the plane
The distance of a point $(x_0,y_0,z_0)$ a plane $ax + by + cz - d = 0$
$\frac {|ax + by + cz - d|}{\sqrt {a^2 + b^2 + c^2}}$
or $d = \frac {(p_1-p_0)\cdot n}{\|n\|}$ where $p_1$ is a point on the plane, and $p_0$ is your point.
and in the specific case $\frac 3{\sqrt 3}$
The reflection of your point. Travel twice the distance along the normal line.
$(1,2,3) - 2 \frac {3}{\sqrt 3}\frac {n}{\|n\|} = (1,2,3) - 2(1,1,1) = -1,0,1$