The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is
\(\cfrac{x-2}{5-2}=\)
Therefore, points on the line
2 is of the form (x, y, z) = (3λ + 2, 4λ − 1, 2λ + 2).
Since, the line intersects the plane x − y + z = 5.
Therefore, for intersection point of the line and plane, the equation of the points on the line must satisfy the equation of the plane.
Therefore, the intersection point of the line and the plane is Q(3(0) + 2, 4(0) − 1, 2(0) + 2) = Q(2, −1, 2).
The distance of the point P(−1, −5, −10) from the point Q(2, −1, 2) is
Hence, the distance of the point p(−1, −5, −10) from the point of intersection of the line joining the points a(2, −1, 2) and B(5, 3, 4) with the plane x − y + z = 5 is 13.
