Given: - Point P(2, 3, 4) and the equation of the 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 = 4 – 2λ, y = 6λ, z = 1 – 3λ

Therefore, coordinates of Q( – 2λ + 4, 6λ, – 3λ + 1)

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 + 4 – 2), (6λ – 3), ( – 3λ + 1 – 4)

= ( – 2λ + 2), (6λ – 3), ( – 3λ – 3)

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

(hint: denominator terms of line equation)

= ( – 2,6, – 3)

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) + (6)(6λ – 3) – 3( – 3λ – 3) = 0

⇒ 4λ – 4 + 36λ – 18 + 9λ + 9 = 0

⇒ 49λ – 13 = 0

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

units