Question
Find the distance of a point P (5, 3, 4) from the point where the line 
meets the plane x + y + z = 2.
Solution
The given line is
...(1)
and the given plane is
x + y + z = 2 ...(2)
Any point on (1) is (r + 3, 2r + 4, 2r + 5)
Let it be the point of intersection of line (1) and plane (2).
∴ point (r + 3, 2r + 4, 2 r + 5) lies on (2)
∴ r + 3 + 2r + 4 + 2r + 5 = 2 ⇒ 5r = – 10 ⇒ r = –2
∴ point of intersection is (1, 0, 1)
∴ required distance = distance between points (5, 3, 4) and (1, 0, 1)
...(1)and the given plane is
x + y + z = 2 ...(2)
Any point on (1) is (r + 3, 2r + 4, 2r + 5)
Let it be the point of intersection of line (1) and plane (2).
∴ point (r + 3, 2r + 4, 2 r + 5) lies on (2)
∴ r + 3 + 2r + 4 + 2r + 5 = 2 ⇒ 5r = – 10 ⇒ r = –2
∴ point of intersection is (1, 0, 1)
∴ required distance = distance between points (5, 3, 4) and (1, 0, 1)
