It is given that,

We have a perpendicular from the point say C (2, 3, -8) to the line, of which equation is,

Or,

So, we can say that the vector equation of line is,

We need to find the foot of the perpendicular from the point C(2, 3, -8) to the given line.

Also, the perpendicular distance from the given point C to the line.

Let us find the point of intersection between the point C(2,3, -8) and the given line.

Take,

From ,

⇒ x – 4 = -2λ

⇒ x = 4 – 2λ

From

⇒ y = 6λ

From

⇒ z – 1 = -3λ

⇒ z = 1 – 3λ

We have,

x = 4 – 2λ, y = 6λ and z = 1 – 3λ.

So, coordinates of any point on the given line is (4 – 2λ, 6λ, 1 – 3λ).

Let the foot of perpendicular from the point C(2, 3, -8) on the line be L(4 – 2λ, 6λ, 1 – 3λ).

We know the direction ratio of any line segment CL, where C(x₁, y₁, z₁) and L(x₂, y₂, z₂), is given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

∴, Direction ratio of is given by

Direction ratio of = (4 – 2λ – 2, 6λ – 3, 1 – 3λ – (-8))

⇒ Direction ratio of = (4 – 2 – 2λ, 6λ – 3, 1 + 8 – 3λ)

⇒ Direction ratio of = (2 – 2λ, 6λ – 3, 9 – 3λ)

Also, direction ratio of the line is,

(-2, 6, -3)

Since, L is the foot of the perpendicular on the line, then

Sum of the product of these direction ratios (2 – 2λ, 6λ – 3, 9 – 3λ) and (-2, 6, -3) is 0.

-2(2 – 2λ) + 6(6λ – 3) + (-3)(9 – 3λ) = 0

⇒ -4 + 4λ + 36λ – 18 – 27 + 9λ = 0

⇒ (4λ + 36λ + 9λ) + (-4 – 18 – 27) = 0

⇒ 49λ – 49 = 0

⇒ 49λ = 49

⇒ λ = 1

Substituting λ = 1 in L(4 – 2λ, 6λ, 1 – 3λ). We get

L(4 – 2λ, 6λ, 1 – 3λ) = L(4 – 2(1), 6(1), 1 – 3(1))

⇒ L(4 – 2λ, 6λ, 1 – 3λ) = L(4 – 2, 6, 1 – 3)

⇒ L(4 – 2λ, 6λ, 1 – 3λ) = L(2, 6, -2)

Now, let us find the perpendicular distance from point C to the line, that is, point L.

⇒ We need to find .

We know that,

Put λ = 1,

To find ,

Thus, the foot of the perpendicular from the point to the given line is (2, 6, -2) and the perpendicular distance is 3√5 units.