The distance between a point $$P$$ and a plane $$\pi$$, $$\text{d}(P,\pi)$$, is the minimal distance between $$P$$ and any point of the plane.

- If $$P$$ is a point of the plane $$\pi$$, the distance is zero.
- If $$P$$ is not a point of the plane $$\pi$$, the distance is the module of $$\overrightarrow{PP'}$$, where $$P'$$ the orthogonal projection of $$P$$ on the plane $$\pi$$.

Nevertheless, there exists a much more practical formula (but not easy to obtain) that is presented next:

Let $$P =(p_1,p_2,p_3)$$ and let $$\pi: Ax+By+Cz+D = 0$$. Then,

$$$\text{d}(P,\pi)=\dfrac{|A\cdot p_1+B\cdot p_2+C\cdot p_3+D|} {\sqrt{A^2+B^2+C^2}}$$$

Calculate the distance between the point $$P=(-2,0,3)$$ and the plane $$\pi:4x+2y-4z+3=0$$.

We can apply the formula: $$$\text{d}(P,\pi)=\dfrac{|A\cdot p_1+B\cdot p_2+C\cdot p_3+D|} {\sqrt{A^2+B^2+C^2}} = \dfrac{|4\cdot(-2)+2\cdot0-4\cdot3+3|}{\sqrt{4^2+2^2+(-4)^2}} = \dfrac{17}{6}$$$