Find the equation of the perpendicular drawn from the point (1,-2,3) to the plane 2x - 3y + 4z +9 = 0. Also find the co-ordinates of the

foot of the perpendicular.

the direction ratios of the line perpendicular to the plane $2x-3y+4z+9=0.....\left(1\right) is 2,-3,4$
equation of a line passing through the point (1,-2,3) and perpendicular to plane (1) is;
$\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}$ ..........(2)
coordinates of any point on the line (2) is given by:
$$\begin{gathered} \frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}=r \\ x=2r+1, y=-3r-2,z=4r+3 \end{gathered}$$
the foot of the perpendicular is the intersection point of the plane (1) and line (2):
therefore
$$\begin{gathered} 2*(2r+1)-3*(-3r-2)+4*(4r+3)+9=0 \\ 4r+2+9r+6+16r+12+9=0 \\ 29r+29=0 \\ r=-1 \end{gathered}$$
thus the coordinates of the foot of perpendicular is $(-2+1,3-2,-4+3) i.e. (-1,1,-1)$

hope this helps you