Given: Line meets plane 2x – y + 3z – 1 = 0

To find: Point of intersection of line and plane.

Explanation:

Let the equation of the line be

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

Since this point also lies on the plane,

2(3λ + 1) - (4λ -2) + 3(-2λ +3) = 1

6λ + 2 – 4λ + 2 – 6λ + 9 = 1

-4λ = -12

λ = 3

Therefore required point is (10, 10, -3)