Given: - Point P(1, 0, 0) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = 2λ + 1, y = – 3λ – 1, z = 8λ – 10

Therefore, coordinates of Q(2λ + 1, – 3λ – 1,8λ – 10)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ + 1 – 1), ( – 3λ – 1 – 0), (8λ – 10 – 0)

= (2λ), ( – 3λ – 1), (8λ – 10)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2, – 3,8)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ) + ( – 3)( – 3λ – 1) + 8(8λ – 10) = 0

⇒ 4λ + 9λ + 3 + 64λ – 80 = 0

⇒ 77λ – 77 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

Distance between PQ

Tip: – Distance between two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

= 2√6 unit