The coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0 are :

$\left(\frac{-3}{61}, \frac{-4}{61}, \frac{6}{61}\right)$

The equation of the plane is 3x + 4y – 6z + 1 = 0           ……(1)

The direction ratios of the normal to the plane (1) are 3, 4, –6. So equation of the line through (0, 0, 0) and perpendicular to the plane (1) are

$\frac{x}{3}=\frac{y}{4}=\frac{z}{-6}$ = r (say)         ……(2)

The coordinates of any point P on (2) are (3r, 4r, –6r). If this point lie on the plane (1), then

3(3r) + 4(4r) – 6(–6r) + 1 = 0 i.e. r = $-\frac{1}{61}$

Putting the value of r coordinates of the foot of the perpendicular P are $\left(\frac{-3}{61}, \frac{-4}{61}, \frac{6}{61}\right)$.