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

Answer

To find the coordinates of the points on the line defined 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 given line can be parameterized using a parameter \(t\). We can express the coordinates of any point on the line as follows:

\[ x = 1 + 2t \]
\[ y = -2 + 3t \]
\[ z = 3 + 6t \]

### Step 2: Set Up the Distance Formula
We need to find the distance between the point on the line \((x, y, z)\) and the point \((1, -2, 3)\). The distance \(d\) can be calculated using the distance formula:

\[ d = \sqrt{(x - 1)^2 + (y + 2)^2 + (z - 3)^2} \]

### Step 3: Substitute the Parameterized Coordinates
Substituting the parameterized coordinates into the distance formula gives:

\[ d = \sqrt{((1 + 2t) - 1)^2 + ((-2 + 3t) + 2)^2 + ((3 + 6t) - 3)^2} \]

This simplifies to:

\[ d = \sqrt{(2t)^2 + (3t)^2 + (6t)^2} \]
\[ = \sqrt{4t^2 + 9t^2 + 36t^2} \]
\[ = \sqrt{49t^2} \]
\[ = 7|t| \]

### Step 4: Set the Distance Equal to 3
We know that the distance \(d\) should equal 3 units:

\[ 7|t| = 3 \]

### Step 5: Solve for \(t\)
Dividing both sides by 7 gives:

\[ |t| = \frac{3}{7} \]

This implies two cases:

1. \(t = \frac{3}{7}\)
2. \(t = -\frac{3}{7}\)

### Step 6: Find the Coordinates for Each \(t\)
Now we will find the coordinates for both values of \(t\).

#### Case 1: \(t = \frac{3}{7}\)

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

So, the coordinates for \(t = \frac{3}{7}\) are:

\[ \left(\frac{13}{7}, -\frac{5}{7}, \frac{39}{7}\right) \]

#### Case 2: \(t = -\frac{3}{7}\)

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

So, the coordinates for \(t = -\frac{3}{7}\) are:

\[ \left(\frac{1}{7}, -\frac{23}{7}, \frac{3}{7}\right) \]

### 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. \(\left(\frac{13}{7}, -\frac{5}{7}, \frac{39}{7}\right)\)
2. \(\left(\frac{1}{7}, -\frac{23}{7}, \frac{3}{7}\right)\)