Let point P = (2, 3, 7) and Q be the foot of the perpendicular drawn from P to the plane 3x – y – z = 7.

Direction ratios of PQ are proportional to 3, –1, –1 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) = (2, 3, 7) and (l, m, n) = (3, –1, –1)

Hence, the equation of PQ is

⇒ x = 3α + 2, y = 3 – α, z = 7 – α

Let Q = (3α + 2, 3 – α, 7 – α).

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

⇒ 3(3α + 2) – (3 – α) – (7 – α) = 7

⇒ 9α + 6 – 3 + α – 7 + α = 7

⇒ 11α – 4 = 7

⇒ 11α = 11

∴ α = 1

We have Q = (3α + 2, 3 – α, 7 – α)

⇒ Q = (3×1 + 2, 3 – 1, 7 – 1)

∴ Q = (5, 2, 6)

Using the distance formula, we have

Thus, the required foot of perpendicular is (5, 2, 6) and the length of the perpendicular is units.