Given: - Point P(0, 2, 7) 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 = – λ – 2, y = 3λ + 1, z = – 2λ + 3

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

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

= ( – λ – 2 – 0), (3λ + 1 – 2), ( – 2λ + 3 – 7)

= ( – λ – 2), (3λ – 1), ( – 2λ – 4)

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

(hint: denominator terms of line equation)

= ( – 1,3, – 2)

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.

⇒ – 1( – λ – 2) + (3)(3λ – 1) – 2( – 2λ – 4) = 0

⇒ λ + 2 + 9λ – 3 + 4λ + 8 = 0

⇒ 14λ + 7 = 0

⇒

Therefore coordinates of Q

i.e. Foot of perpendicular

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