Find the coordinates of the points on the line `(x +1)/(2) = (y - 2)/(3) = (z + 3)/(6)`, which are at a distance of 3 units from the point ( -1 , 2 , -3).

Answer

To find the coordinates of the points on the line given by the equation \((x + 1)/2 = (y - 2)/3 = (z + 3)/6\) that are at a distance of 3 units from the point \((-1, 2, -3)\), we can follow these steps:

### Step 1: Parameterize the Line
The equation of the line can be parameterized using a parameter \(k\). We can express the coordinates \(x\), \(y\), and \(z\) in terms of \(k\):
- From \((x + 1)/2 = k\), we get:
\[ x = 2k - 1 \]
- From \((y - 2)/3 = k\), we get:
\[ y = 3k + 2 \]
- From \((z + 3)/6 = k\), we get:
\[ z = 6k - 3 \]

### Step 2: Write the Distance Formula
We need to find the distance between the point \((-1, 2, -3)\) and the point \((2k - 1, 3k + 2, 6k - 3)\). The distance \(d\) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(2k - 1 + 1)^2 + (3k + 2 - 2)^2 + (6k - 3 + 3)^2} \]
This simplifies to:
\[ d = \sqrt{(2k)^2 + (3k)^2 + (6k)^2} \]

### Step 3: Set the Distance Equal to 3
We set the distance equal to 3:
\[ \sqrt{(2k)^2 + (3k)^2 + (6k)^2} = 3 \]
Squaring both sides gives:
\[ (2k)^2 + (3k)^2 + (6k)^2 = 9 \]
This simplifies to:
\[ 4k^2 + 9k^2 + 36k^2 = 9 \]
\[ 49k^2 = 9 \]

### Step 4: Solve for \(k\)
Now, we solve for \(k\):
\[ k^2 = \frac{9}{49} \]
Taking the square root gives:
\[ k = \pm \frac{3}{7} \]

### Step 5: Find the Coordinates for Each Value of \(k\)
Now we substitute \(k\) back into the parameterized equations to find the coordinates:

1. For \(k = \frac{3}{7}\):
- \(x = 2\left(\frac{3}{7}\right) - 1 = \frac{6}{7} - 1 = -\frac{1}{7}\)
- \(y = 3\left(\frac{3}{7}\right) + 2 = \frac{9}{7} + 2 = \frac{9}{7} + \frac{14}{7} = \frac{23}{7}\)
- \(z = 6\left(\frac{3}{7}\right) - 3 = \frac{18}{7} - 3 = \frac{18}{7} - \frac{21}{7} = -\frac{3}{7}\)

So, one point is \(\left(-\frac{1}{7}, \frac{23}{7}, -\frac{3}{7}\right)\).

2. For \(k = -\frac{3}{7}\):
- \(x = 2\left(-\frac{3}{7}\right) - 1 = -\frac{6}{7} - 1 = -\frac{6}{7} - \frac{7}{7} = -\frac{13}{7}\)
- \(y = 3\left(-\frac{3}{7}\right) + 2 = -\frac{9}{7} + 2 = -\frac{9}{7} + \frac{14}{7} = \frac{5}{7}\)
- \(z = 6\left(-\frac{3}{7}\right) - 3 = -\frac{18}{7} - 3 = -\frac{18}{7} - \frac{21}{7} = -\frac{39}{7}\)

So, the second point is \(\left(-\frac{13}{7}, \frac{5}{7}, -\frac{39}{7}\right)\).

### Final Answer
The coordinates of the points on the line that are at a distance of 3 units from the point \((-1, 2, -3)\) are:
1. \(\left(-\frac{1}{7}, \frac{23}{7}, -\frac{3}{7}\right)\)
2. \(\left(-\frac{13}{7}, \frac{5}{7}, -\frac{39}{7}\right)\)