Let point P = (0, 0, 0) and Q be the foot of the perpendicular drawn from P to the plane 2x – 3y + 4z – 6 = 0.

Direction ratios of PQ are proportional to 2, –3, 4 as PQ is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (0, 0, 0) and (l, m, n) = (2, –3, 4)

Hence, the equation of PQ is

⇒ x = 2α, y = –3α, z = 4α

Let Q = (2α, –3α, 4α).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ 2(2α) – 3(–3α) + 4(4α) – 6 = 0

⇒ 4α + 9α + 16α – 6 = 0

⇒ 29α = 6

We have Q = (2α, –3α, 4α)

Thus, the required foot of perpendicular is.