Let point P = (1, 1, 2) and Q be the foot of the perpendicular drawn from to P the plane.

Direction ratios of PQ are proportional to 1, –2, 4 as PQ is normal to the plane and parallel to.

Recall the vector equation of the line passing through the point with position vector and parallel to vector is given by

Here, and

Hence, the equation of PQ is

Let the position vector of Q be. As Q is a point on this line, for some scalar α, we have

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

⇒ (1 + α)(1) + (1 – 2α)(–2) + (2 + 4α)(4) = –5

⇒ 1 + α – 2 + 4α + 8 + 16α = –5

⇒ 21α + 7 = –5

⇒ 21α = –12

Foot of the perpendicular

Thus,

Using the distance formula, we have

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