Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7.

The equation of a line passing through two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is

⇒

Given the line passes through the points A(3, –4, –5) and

B(2, –3, 1)

∴ x₁ = 3, y₁ = -4, z₁ = -5

and, x₂ = 2, y₂ = -3, z₂ = 1

So, the equation of line is

⇒

⇒

So,

x = -k + 3 | y = k - 4 | z = 6k - 5 …(1)

Let (x, y, z) be the coordinates of the point where the line crosses the plane 2x + y + z + 7 = 0

Putting the value of x,y,z from (1) in the equation of plane,

2x + y + z + 7 = 0

⇒ 2(-k + 3) + (k - 4) + (6k - 5) = 7

⇒ 5k - 3 = 7

⇒ 5k = 10

∴ k = 2

Putting the value of k in x, y, z

x = - k + 3 = - 2 + 3 = 1

y = k - 4 = 2 - 4 = - 2

z = 6k - 5 = 12 - 5 = 7

Hence, the coordinates of the required point are (1, -2,7).