Answer :

Given -

The equation of line is



and the equation of the plane is



To find the intersection of line and plane, putting value of from equation of line into equation of plane, we get -




(2 + 3λ) × 1 + (-1 + 4λ) × (-1) + (2 + 2λ) × 1 = 5


2 + 3λ + 1 - 4λ + 2 + 2λ = 5


λ = 0


So, the equation of line is



Let the point of intersection be (x,y,z)


So,



Hence, x = 2, y = -1, z = 2


Therefore, the point of intersection is (2, -1, 2).


Now, the distance between points (x1, y1, z1) and (x2, y2, z2) is given by -


units


Distance between the points A(2, -1, 2) and B(-1, -5, -10) is given by -






= 13 units


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