To find the coordinates of the points on the line given by \((x-1)/2 = (y+2)/3 = (z-3)/6\) that are at a distance of 3 units from the point \((1, -2, 3)\), we can follow these steps:
### Step 1: Parametrize the line
The line can be expressed in parametric form. Let \( k \) be the parameter. We can write:
\[
x = 2k + 1, \quad y = 3k - 2, \quad z = 6k + 3
\]
### Step 2: Set up the distance formula
We need to find the distance from the point \((1, -2, 3)\) to a point \((x, y, z)\) on the line. The distance \(d\) between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
In our case, we want this distance to be 3 units:
\[
\sqrt{(x - 1)^2 + (y + 2)^2 + (z - 3)^2} = 3
\]
### Step 3: Substitute the parametric equations into the distance formula
Substituting the parametric equations into the distance formula gives:
\[
\sqrt{((2k + 1) - 1)^2 + ((3k - 2) + 2)^2 + ((6k + 3) - 3)^2} = 3
\]
This simplifies to:
\[
\sqrt{(2k)^2 + (3k)^2 + (6k)^2} = 3
\]
### Step 4: Square both sides to eliminate the square root
Squaring both sides, we get:
\[
(2k)^2 + (3k)^2 + (6k)^2 = 9
\]
This simplifies to:
\[
4k^2 + 9k^2 + 36k^2 = 9
\]
\[
49k^2 = 9
\]
### Step 5: Solve for \(k\)
Now, we can solve for \(k\):
\[
k^2 = \frac{9}{49} \implies k = \pm \frac{3}{7}
\]
### Step 6: Find the coordinates for both values of \(k\)
Substituting \(k = \frac{3}{7}\):
1. For \(k = \frac{3}{7}\):
\[
x = 2\left(\frac{3}{7}\right) + 1 = \frac{6}{7} + 1 = \frac{13}{7}
\]
\[
y = 3\left(\frac{3}{7}\right) - 2 = \frac{9}{7} - 2 = \frac{9}{7} - \frac{14}{7} = -\frac{5}{7}
\]
\[
z = 6\left(\frac{3}{7}\right) + 3 = \frac{18}{7} + 3 = \frac{18}{7} + \frac{21}{7} = \frac{39}{7}
\]
2. For \(k = -\frac{3}{7}\):
\[
x = 2\left(-\frac{3}{7}\right) + 1 = -\frac{6}{7} + 1 = \frac{1}{7}
\]
\[
y = 3\left(-\frac{3}{7}\right) - 2 = -\frac{9}{7} - 2 = -\frac{9}{7} - \frac{14}{7} = -\frac{23}{7}
\]
\[
z = 6\left(-\frac{3}{7}\right) + 3 = -\frac{18}{7} + 3 = -\frac{18}{7} + \frac{21}{7} = \frac{3}{7}
\]
### Final Result
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. \(\left(\frac{1}{7}, -\frac{23}{7}, \frac{3}{7}\right)\)
