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 coordinate of a point B of line L such that AB is parallel to the plane is

Options

A. (a) `(10, -1, 15)`
B. (b) `(-5, 4, -5)`
C. (c) `(4, 1, 7)`
D. (d) `(-8, 5, -9)`

Explanation

To find the coordinates of point B on line L such that line segment AB is parallel to the plane given by the equation \(x + y - z = 1\), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as:
\[ x + y - z = 1 \]
The normal vector \( \mathbf{n} \) to the plane can be derived from the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n} = (1, 1, -1) \]

### Step 2: Write the parametric equations of the line L
The line L is given by the equations:
\[ x = 1 + 3r, \quad y = 2 - r, \quad z = 3 + 4r \]
We can express this in vector form as:
\[ \mathbf{L}(r) = (1 + 3r, 2 - r, 3 + 4r) \]

### Step 3: Find the vector AB
Let point A be \(A(1, 2, -3)\) and point B be represented as \(B(1 + 3r, 2 - r, 3 + 4r)\). The vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (1 + 3r - 1, 2 - r - 2, 3 + 4r + 3) = (3r, -r, 6 + 4r) \]

### Step 4: Set up the condition for parallelism
For the line segment AB to be parallel to the plane, the vector \( \overrightarrow{AB} \) must be perpendicular to the normal vector \( \mathbf{n} \). This means:
\[ \overrightarrow{AB} \cdot \mathbf{n} = 0 \]
Calculating the dot product:
\[ (3r, -r, 6 + 4r) \cdot (1, 1, -1) = 3r - r - (6 + 4r) = 0 \]
Simplifying this gives:
\[ 3r - r - 6 - 4r = 0 \implies -2r - 6 = 0 \implies -2r = 6 \implies r = -3 \]

### Step 5: Substitute r back to find point B
Now that we have \(r = -3\), we can substitute this value back into the parametric equations of the line L to find the coordinates of point B:
\[ x = 1 + 3(-3) = 1 - 9 = -8 \]
\[ y = 2 - (-3) = 2 + 3 = 5 \]
\[ z = 3 + 4(-3) = 3 - 12 = -9 \]
Thus, the coordinates of point B are:
\[ B(-8, 5, -9) \]

### Final Answer
The coordinates of point B such that AB is parallel to the plane are:
\[ \boxed{(-8, 5, -9)} \]