Find the distance

Given line,

Let,

⇒ x = 3k+2 , y = 4k-1 ,z = 12k+2

Coordinates of any point on the line are (3k+2 , 4k-1 , 12k+2).

The point of intersection of the line and the plane x – y + z = 5 will be in the form (3k+2 , 4k-1 , 12k+2).

we have, plane x – y + z = 5

⇒ 3k+2 – (4k-1)+ 12k+2 =5

⇒ 3k+2 – 4k+1+ 12k+2-5 =0

⇒ 11k = 0

⇒ k = 0

Hence, the point of intersection of the line and the plane x – y + z = 5 is (2 , -1 , 2).

The required distance between the (-1, -5, -10) and (2 , -1 , 2)

=

=

=13 units