Let point P = (1, 3/2, 2) and Q be the foot of the perpendicular drawn from P to the plane 2x – 2y + 4z + 5 = 0.

Direction ratios of PQ are proportional to 2, –2, 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) = (1,, 2) and (l, m, n) = (2, –2, 4)

Hence, the equation of PQ is

Let.

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

⇒ 4α + 2 – (–4α + 3) + 16α + 8 + 5 = 0

⇒ 20α + 4α – 3 + 15 = 0

⇒ 24α = –12

We have

Using the distance formula, we have

Thus, the required foot of perpendicular is and the length of the perpendicular is units.