Given: Line meets plane 2x + 3y – z = 14

To find: Point of intersection of line and plane.

Explanation:

Let the equation of the line be

Therefore, any point on the line is (2λ + 1, 4λ +2, -3λ +3)

Since this point also lies on the plane,

2(2λ + 1) + 3(4λ +2) – (-3λ +3) =14

4λ + 2 + 12λ + 6 + 3λ – 3 = 14

19λ + 5 = 14

Therefore the required point is (3, 5, 7).