Given: – Point P(1, 2, – 3) 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 = – 2λ + 3, z = – λ

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

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), ( – 2λ + 3 – 2), ( – λ + 3)

= (2λ – 2), ( – 2λ + 1), ( – λ + 3)

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

(hint: denominator terms of line equation)

= (2, – 2, – 1)

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 – 2) + ( – 2)( – 2λ + 1) – 1( – λ + 3) = 0

⇒ 4λ – 4 + 4λ – 2 + λ – 3 = 0

⇒ 9λ – 9 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

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