We know that the equation passing through two point (a, b, c) and (d, e, f) is given by

the line through A(3, – 4, – 5) and B(2, – 3, 1) is

Now let us see how a point is going to be on the line

X = – k + 3, y = k – 4, z = 6k – 5

So now let a point P be the point of the intersection of a line and the plane so let the coordinates of P = (– k + 3, k – 4, 6k – 5)

Now let us find the equation of the plane passing through L(2, 2, 1), M(3, 0, 1) and N(4, – 1, 0).

The equation of the plane passing through these three points is given by the following equation.

(x – 2)2 – (y – 2)(– 1) + (z – 1)(– 3 + 4) = 0

2x – 4 + y – 2 – z + 1 = 0

2x + y – z = 5

Now point P lies on this plane so

2(3 – k) + (k – 4) – (6k – 5) = 5

6 – 2k + k – 4 – 6k + 5 = 5

– 7k = – 2

So point P is

Now we have to find the ration in which this P divides AB

Let the ratio b m:1

We know the section formula

That is if a line AB is divided by P in ratio m:1 then

Solving

We get 19m + 19 = 14m + 21

5m = – 2

So point P divides line in ration 2:5 externally since ratio is negative.