Q.If the image of the point $A(1, 2, 3)$ with respect to the plane $\vec{r}\cdot\left(\hat{i}+2\hat{j}+4\hat{k}\right)=38$ is $A'(3, 6, 11)$, then the foot of the perpendicular from the point A to the given plane is :

Four closely related "how far / how tilted" computations, all built from a plane's normal $\vec n$ or a line's direction $\vec b$.

Angle between two planes = angle between their normals: $\theta=\cos^{-1}\!\left(\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}\right)$; perpendicular iff $\vec n_1\cdot\vec n_2=0$, parallel iff $\vec n_1=\lambda\vec n_2$.

Angle between a line and a plane = complement of the angle between the line's direction $\vec b$ and the plane's normal $\vec n$ (since a line lying flat in the plane is perpendicular to the normal, and vice versa): $\theta=\sin^{-1}\!\left(\left|\dfrac{\vec b\cdot\vec n}{|\vec b||\vec n|}\right|\right)$; the line is perpendicular to the plane iff $\vec b\parallel\vec n$, and parallel to the plane iff $\vec b\cdot\vec n=0$.

Distance from a point $\vec u$ to a plane $\vec r\cdot\vec n=p$: $\delta=\dfrac{|\vec u\cdot\vec n-p|}{|\vec n|}$ (Cartesian: $\delta=\dfrac{|ax_1+by_1+cz_1-p|}{\sqrt{a^2+b^2+c^2}}$); taking $\vec u=\vec 0$ gives the distance from the origin, $\delta=\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}$ for $ax+by+cz+d=0$. The foot of that perpendicular is $\vec u+\dfrac{p-\vec u\cdot\vec n}{|\vec n|^2}\vec n$. …