To solve the problem, we need to find the coordinates of a point on the line given by the equation
\[
\frac{x-1}{2} = \frac{y+1}{-3} = z
\]
that is at a distance of \(4\sqrt{14}\) from the point \((1, -1, 0)\).
### Step 1: Parameterize the line
We can express the coordinates of any point on the line in terms of a parameter \( \lambda \).
From the equation of the line, we can write:
\[
x = 2\lambda + 1
\]
\[
y = -3\lambda - 1
\]
\[
z = \lambda
\]
### Step 2: Write the distance formula
The distance \(d\) between the point \((x, y, z)\) on the line and the point \((1, -1, 0)\) is given by:
\[
d = \sqrt{(x - 1)^2 + (y + 1)^2 + (z - 0)^2}
\]
Substituting the expressions for \(x\), \(y\), and \(z\):
\[
d = \sqrt{((2\lambda + 1) - 1)^2 + ((-3\lambda - 1) + 1)^2 + (\lambda - 0)^2}
\]
This simplifies to:
\[
d = \sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2}
\]
### Step 3: Set the distance equal to \(4\sqrt{14}\)
We know the distance is \(4\sqrt{14}\), so we set up the equation:
\[
\sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2} = 4\sqrt{14}
\]
Squaring both sides gives:
\[
(2\lambda)^2 + (-3\lambda)^2 + \lambda^2 = (4\sqrt{14})^2
\]
This simplifies to:
\[
4\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \times 14
\]
\[
14\lambda^2 = 224
\]
### Step 4: Solve for \(\lambda\)
Dividing both sides by 14:
\[
\lambda^2 = 16
\]
Taking the square root:
\[
\lambda = 4 \quad \text{or} \quad \lambda = -4
\]
### Step 5: Find the coordinates for both values of \(\lambda\)
1. For \(\lambda = 4\):
\[
x = 2(4) + 1 = 9
\]
\[
y = -3(4) - 1 = -13
\]
\[
z = 4
\]
Thus, one point is \((9, -13, 4)\).
2. For \(\lambda = -4\):
\[
x = 2(-4) + 1 = -7
\]
\[
y = -3(-4) - 1 = 11
\]
\[
z = -4
\]
Thus, the other point is \((-7, 11, -4)\).
### Final Answer
The coordinates of the points on the line at a distance \(4\sqrt{14}\) from the point \((1, -1, 0)\) are:
\[
(9, -13, 4) \quad \text{and} \quad (-7, 11, -4)
\]
