The perpendicular distance from the origin to point in the plane The plane $3x-2y-z=-4$ is passing through $A(1,2,3)$ and parallel to $u=2i+3j$ and $v=i+2j-k$.
The perpendicular distance from the origin to the plane is $r.n = d$ but how to determine the point (Call it N) on the plane and what's the coordinate of the point N?
 A: A point on the normal to the plane passing through the origin has coordinates $(3t,-2t,-t)$ [because $(3,-2,-1)$ is normal to the plane] and if it is on the plane $3x-2y-z=-4$ we have $9t+4t+t=-4$ so $t=-\dfrac{2}{7}$ and the point is $(-6/7,4/7,2/7)$.
A: There is a general procedure & formula derived in Reflection formula by HCR to calculate the point of reflection $\color{blue}{P'(x', y', z')}$ of the any point $\color{blue}{P(x_{o}, y_{o}, z_{o})}$ about the plane: $\color{blue}{ax+by+cz+d=0}$ & hence the foot of perpendicular say point $N$ is determined as follows 
$$\color{blue}{N\equiv\left(\frac{x_{o}+x'}{2}, \frac{y_{o}+y'}{2}, \frac{z_{o}+z'}{2}\right)}$$ Where $$\color{red}{x'=x_{o}-\frac{2a(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$   $$\color{red}{y'=y_{o}-\frac{2b(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$   $$\color{red}{z'=z_{o}-\frac{2c(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$
As per your question, the foot of perpendicular $N$ drawn from the origin $\color{blue}{(0, 0, 0)\equiv(x_{o}, y_{o}, z_{o})}$ to the given plane: $\color{blue}{3x-2y-z+4=0}$ is determined by setting the corresponding values in the above expression as follows $$\color{}{x'=0-\frac{2(3)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=-\frac{24}{14}=-\frac{12}{7}$$ $$\color{}{y'=0-\frac{2(-2)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=\frac{16}{14}=\frac{8}{7}$$   $$\color{black}{z'=0-\frac{2(-1)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=\frac{8}{14}=\frac{4}{7}$$ Now, setting these values, we get foot of perpendicular $N$ $$N\equiv\left(\frac{0+\left(-\frac{12}{7}\right)}{2}, \frac{0+\frac{8}{7}}{2}, \frac{0+\frac{4}{7}}{2}\right)$$ $$\color{blue}{N\equiv\left(-\frac{6}{7}, \frac{4}{7}, \frac{2}{7}\right)}$$   
