Find the distance of the point (–1, –5, –10) from the point of intersection of the line and the plane .

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 (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is given by -

units

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

= 13 units

18