Determine the coordinates of the foot of the perpendicular drawn from the origin to the plane $4x-2y+3z-6=0$.

The plane is $4x-2y+3z=6$.The normal to the plane has direction ratios $(4,-2,3)$.The equation of the line passing through the origin $(0,0,0)$ and perpendicular to the plane is:
$$\frac{x}{4}=\frac{y}{-2}=\frac{z}{3}=\lambda$$
Any point on this line is $(4\lambda ,-2\lambda ,3\lambda )$.This point lies on the plane, so:
$$4(4\lambda )-2(-2\lambda )+3(3\lambda )=6$$
$$16\lambda +4\lambda +9\lambda =6$$
$$29\lambda =6⟹\lambda =\frac{6}{29}$$
Substituting $\lambda$:
$$x=4\left(\frac{6}{29}\right)=\frac{24}{29}$$
$$y=-2\left(\frac{6}{29}\right)=-\frac{12}{29}$$
$$z=3\left(\frac{6}{29}\right)=\frac{18}{29}$$
The foot of the perpendicular is $\left(\frac{24}{29},-\frac{12}{29},\frac{18}{29}\right)$.