The coordinates of a point on the line `(x-1)/(2)=(y+1)/(-3)=z` at a distance `4sqrt(14)` from the point `(1, -1, 0)` are

Options

A. `(9, -13, 4)`
B. `(8sqrt(14)+1, -12sqrt(14)-1, 4sqrt(14))`
C. `(-7, 11, -4)`
D. `(-8sqrt(14)+1, 12sqrt(14)-1, -4sqrt(14))`

Answer

To find the coordinates of a point on the line given by the equation \((x-1)/2 = (y+1)/(-3) = z\) that is at a distance of \(4\sqrt{14}\) from the point \((1, -1, 0)\), we can follow these steps:

### Step 1: Parametrize the Line
We start by letting the common ratio equal to a parameter \(\lambda\):
\[ \frac{x-1}{2} = \frac{y+1}{-3} = z = \lambda \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(\lambda\):
\[ x = 2\lambda + 1 \]
\[ y = -3\lambda - 1 \]
\[ z = \lambda \]

### Step 2: Set Up the Distance Formula
Next, we need to find the distance from the point \((1, -1, 0)\) to the point \((x, y, z)\) we just defined. The distance \(d\) can be expressed using the distance formula:
\[ d = \sqrt{(x - 1)^2 + (y + 1)^2 + (z - 0)^2} \]
We know that this distance is equal to \(4\sqrt{14}\):
\[ \sqrt{(x - 1)^2 + (y + 1)^2 + z^2} = 4\sqrt{14} \]

### Step 3: Substitute the Parametric Equations
Substituting the expressions for \(x\), \(y\), and \(z\) into the distance formula gives:
\[ \sqrt{((2\lambda + 1) - 1)^2 + ((-3\lambda - 1) + 1)^2 + (\lambda - 0)^2} = 4\sqrt{14} \]
This simplifies to:
\[ \sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2} = 4\sqrt{14} \]

### Step 4: Simplify the Equation
Now, squaring both sides to eliminate the square root:
\[ (2\lambda)^2 + (-3\lambda)^2 + \lambda^2 = (4\sqrt{14})^2 \]
This simplifies to:
\[ 4\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \cdot 14 \]
\[ 14\lambda^2 = 224 \]
\[ \lambda^2 = 16 \]
\[ \lambda = 4 \quad \text{or} \quad \lambda = -4 \]

### Step 5: Find the Coordinates for Each \(\lambda\)
Now we substitute \(\lambda = 4\) and \(\lambda = -4\) back into the parametric equations to find the coordinates.

1. For \(\lambda = 4\):
\[ x = 2(4) + 1 = 9 \]
\[ y = -3(4) - 1 = -13 \]
\[ z = 4 \]
Thus, the point is \((9, -13, 4)\).

2. For \(\lambda = -4\):
\[ x = 2(-4) + 1 = -7 \]
\[ y = -3(-4) - 1 = 11 \]
\[ z = -4 \]
Thus, the point is \((-7, 11, -4)\).

### Final Answer
The coordinates of the points on the line at a distance of \(4\sqrt{14}\) from the point \((1, -1, 0)\) are:
\[ (9, -13, 4) \quad \text{and} \quad (-7, 11, -4) \]