Answer :

**Given:** A plane passing through L(2, 2, 1), M(3, 0, 1) and N(4, -1, 0).

**To find:** coordinate of the point P where the line through A(3, -4, -5) and B(2, -3, 1) crosses the plane

The equation of plane crossing through these points is

Applying R_{2} → -R_{2} + R_{1} and R_{3} → -R_{3} + R_{1}

Expanding the determinant,

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

⇒ 2x – 4 + y – 2 + z – 1 = 0

⇒ 2x + y + z – 7 = 0

Therefore, the equation of plane is 2x + y + z – 7 = 0

The equation of the line passing through the point (a, b, c) and (m, n, o) is given by:

The equation of the line passing through point A(3, -4, -5) and B(2, -3, 1) is given by:

This point P lies on both line and plane

Therefore,

2(k + 2) + (-k – 3) – 6k + 1 – 7 = 0

⇒ 2k + 4 – k – 3 – 6k + 1 – 7 = 0

⇒– 5k – 5 = 0

⇒ 5k = – 5

⇒ k = – 1

Hence, Intersection Point P(1, -2, 7)

**Section formula:**

Let P divide the line AB in the ration k:1 and x_{1} = 3, x_{2} = 2

So,

⇒ k + 1 = 2k + 3

⇒ k = -2

Hence the ratio 2:1 and negative sign shows that P is dividing AB externally

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