Find the distance between the point (–1, –5, –10) and the point of intersection of the and the plane

x – y + z = 5.

Given the equation of a line:

∴ x = 3λ + 2 ; y = 4λ – 1 and z = 12λ + 2 where λ is any real number.

∴ (3λ + 2 , 4λ – 1 , 12λ + 2) gives coordinate of a random point which lies on the given line.

At the point of intersection of plane and line, points will be lying on plane too.

So they will satisfy the equation of plane: x – y + z = 5

∴ 3λ + 2 – (4λ - 1) + (12λ + 2) = 5

⇒ 11λ + 5 = 5

∴ λ = 0

Thus point of intersection is (2,-1,2)

But we need to find the distance between (-1,-5,-10) and (2,-1,2)

Distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by-

D =

∴ D =

⇒ D =