Find the co-ordinates of the point P, where the line through A (3,-4,-5) and B(2,-3,1) crosses the plane passing through three points L (2,2,1) , M(3,0,1) and N(4,-1,0). Also , find the ratio in which P divides the line segment AB.

Answer

To find the coordinates of the point P where the line through points A(3, -4, -5) and B(2, -3, 1) intersects the plane defined by the points L(2, 2, 1), M(3, 0, 1), and N(4, -1, 0), we will follow these steps:

### Step 1: Determine the direction vector of line AB
The direction vector of the line AB can be calculated as:
\[ \vec{AB} = B - A = (2 - 3, -3 + 4, 1 + 5) = (-1, 1, 6) \]

### Step 2: Parametric equations of line AB
Using point A and the direction vector, we can write the parametric equations for the line:
\[ x = 3 - t, \quad y = -4 + t, \quad z = -5 + 6t \]
where \( t \) is a parameter.

### Step 3: Find the equation of the plane through points L, M, and N
To find the equation of the plane, we first need to find two vectors in the plane:
\[ \vec{LM} = M - L = (3 - 2, 0 - 2, 1 - 1) = (1, -2, 0) \]
\[ \vec{LN} = N - L = (4 - 2, -1 - 2, 0 - 1) = (2, -3, -1) \]

Next, we find the normal vector to the plane by taking the cross product of \(\vec{LM}\) and \(\vec{LN}\):
\[ \vec{n} = \vec{LM} \times \vec{LN} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{n} = \hat{i}((-2)(-1) - (0)(-3)) - \hat{j}((1)(-1) - (0)(2)) + \hat{k}((1)(-3) - (-2)(2)) \]
\[ = \hat{i}(2) - \hat{j}(-1) + \hat{k}(-3 + 4) = (2, 1, 1) \]

### Step 4: Equation of the plane
Using the normal vector \((2, 1, 1)\) and point L(2, 2, 1), the equation of the plane can be derived as:
\[ 2(x - 2) + 1(y - 2) + 1(z - 1) = 0 \]
Expanding this gives:
\[ 2x + y + z - 7 = 0 \quad \text{or} \quad 2x + y + z = 7 \]

### Step 5: Substitute parametric equations into the plane equation
Substituting the parametric equations of line AB into the plane equation:
\[ 2(3 - t) + (-4 + t) + (-5 + 6t) = 7 \]
Expanding and simplifying:
\[ 6 - 2t - 4 + t - 5 + 6t = 7 \]
\[ (6 - 4 - 5) + (6t - 2t + t) = 7 \]
\[ -3 + 5t = 7 \]
\[ 5t = 10 \quad \Rightarrow \quad t = 2 \]

### Step 6: Find coordinates of point P
Substituting \( t = 2 \) back into the parametric equations:
\[ x = 3 - 2 = 1, \quad y = -4 + 2 = -2, \quad z = -5 + 12 = 7 \]
Thus, the coordinates of point P are:
\[ P(1, -2, 7) \]

### Step 7: Find the ratio in which P divides AB
Using the section formula, if P divides AB in the ratio \( m:n \), then:
\[ \frac{m}{n} = \frac{AP}{PB} \]
Where:
\[ AP = \sqrt{(1 - 3)^2 + (-2 + 4)^2 + (7 + 5)^2} = \sqrt{4 + 4 + 144} = \sqrt{152} = 2\sqrt{38} \]
\[ PB = \sqrt{(1 - 2)^2 + (-2 + 3)^2 + (7 - 1)^2} = \sqrt{1 + 1 + 36} = \sqrt{38} \]
Thus, the ratio \( \frac{AP}{PB} = \frac{2\sqrt{38}}{\sqrt{38}} = 2:1 \).

### Final Answer
The coordinates of point P are \( (1, -2, 7) \) and it divides the line segment AB in the ratio \( 2:1 \) externally.