Consider a plane `x+y-z=1` and point `A(1, 2, -3)`. A line L has the equation `x=1 + 3r, y =2 -r and z=3+4r`. The equation of the plane containing line L and point A has the equation

Options

A. `x-3y+5=0`
B. `x+3y-7=0`
C. `3x-y-1=0`
D. `3x+y-5=0`

Explanation

To find the equation of the plane containing the line \( L \) and the point \( A(1, 2, -3) \), we can follow these steps:

### Step 1: Identify the direction ratios of the line \( L \)
The line \( L \) is given by the parametric equations:
\[ x = 1 + 3r, \quad y = 2 - r, \quad z = 3 + 4r \]
From these equations, we can extract the direction ratios of the line:
- Direction ratio in \( x \): \( 3 \)
- Direction ratio in \( y \): \( -1 \)
- Direction ratio in \( z \): \( 4 \)

Thus, the direction ratios of the line \( L \) are \( (3, -1, 4) \).

### Step 2: Find a normal vector to the plane
The plane must contain the line \( L \) and the point \( A(1, 2, -3) \). To find the normal vector of the plane, we can use the direction ratios of the line and the vector from point \( A \) to any point on the line \( L \).

Let's find a point on the line \( L \) when \( r = 0 \):
\[ x = 1 + 3(0) = 1, \quad y = 2 - 0 = 2, \quad z = 3 + 4(0) = 3 \]
This gives us the point \( B(1, 2, 3) \).

Now, we can find the vector \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B - A = (1 - 1, 2 - 2, 3 - (-3)) = (0, 0, 6) \]

### Step 3: Find the normal vector using cross product
Now, we have two vectors:
1. Direction vector of line \( L \): \( \mathbf{d} = (3, -1, 4) \)
2. Vector \( \overrightarrow{AB} = (0, 0, 6) \)

The normal vector \( \mathbf{n} \) to the plane can be found using the cross product \( \mathbf{d} \times \overrightarrow{AB} \):
\[ \mathbf{n} = (3, -1, 4) \times (0, 0, 6) \]

Calculating the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 4 \\ 0 & 0 & 6 \end{vmatrix} = \mathbf{i}((-1)(6) - (4)(0)) - \mathbf{j}((3)(6) - (4)(0)) + \mathbf{k}((3)(0) - (-1)(0)) \]
\[ = -6\mathbf{i} - 18\mathbf{j} + 0\mathbf{k} = (-6, -18, 0) \]
We can simplify this to \( (1, 3, 0) \) by dividing by -6.

### Step 4: Write the equation of the plane
The general equation of a plane with normal vector \( (a, b, c) \) passing through point \( (x_0, y_0, z_0) \) is given by:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Substituting \( (1, 3, 0) \) for \( (a, b, c) \) and \( (1, 2, -3) \) for \( (x_0, y_0, z_0) \):
\[ 1(x - 1) + 3(y - 2) + 0(z + 3) = 0 \]
This simplifies to:
\[ x - 1 + 3y - 6 = 0 \]
\[ x + 3y - 7 = 0 \]

### Final Answer
The equation of the plane containing line \( L \) and point \( A(1, 2, -3) \) is:
\[ x + 3y - 7 = 0 \]