Find the co-ordinates of the point where the line through the points (3, –4, –5) and (2, –3, 1) crosses the plane 3x + 2y + z + 14 = 0.

Given: line passing through the points (3,–4, –5) and (2,–3, 1), and a plane 3x + 2y + z + 14 = 0.

To find: the coordinates of the point where
the given line crosses the given plane

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

So equation of line passing through two points A(3,–4, –5) and B(2,–3, 1) is

This is equal to constant, so

From this we get

Now let (x,y,z) be the coordinates of the point where the line crosses the plane 3x + 2y + z + 14 = 0.

Putting the x, y, z values from equation (i), (ii) and (iii), in the equation of the given plane, we get

3x + 2y + z + 14 = 0.

⇒ 3(3-k)+2(k-4)+(6k-5)+14=0

⇒ 9-3k+2k-8+6k-5+14=0

⇒ 5k+10=0

⇒ 5k=-10

∴, k=-2

Now substituting the value of k in equation (i), we get

x=3-k⇒ x=3-(-2)⇒ x=5

Now substituting the value of k in equation (ii), we get

y=k-4⇒ y=(-2)-4⇒ y=-6

Now substituting the value of k in equation (iii), we get

z=6k-5⇒ z=6(-2)-5⇒ z=-12-5⇒ z=-17

Hence (1,-2,7) is the coordinates of the point where
the line through the points (3,–4, –5) and (2,–3, 1) crosses the plane 3x + 2y + z + 14 = 0.