Find the co-ordinates of the point at a distance of `sqrt(5)`units from the point `(1,2,3)` on the line `(x+2)/(3) = (y+1)/(2) = (z-3)/(2)`.

Answer

To find the coordinates of the point \( Q \) that is at a distance of \( \sqrt{5} \) units from the point \( P(1, 2, 3) \) on the line given by the equations \( \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2} \), we can follow these steps:

### Step 1: Parameterize the Line
The line can be parameterized using a parameter \( \lambda \). We can express the coordinates of point \( Q \) in terms of \( \lambda \):
\[ \frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = \lambda \]
From this, we can derive the coordinates of \( Q \):
\[ x = 3\lambda - 2, \quad y = 2\lambda - 1, \quad z = 2\lambda + 3 \]

### Step 2: Use the Distance Formula
The distance \( PQ \) between points \( P(1, 2, 3) \) and \( Q(3\lambda - 2, 2\lambda - 1, 2\lambda + 3) \) is given by the formula:
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates, we have:
\[ PQ = \sqrt{(3\lambda - 2 - 1)^2 + (2\lambda - 1 - 2)^2 + (2\lambda + 3 - 3)^2} \]
This simplifies to:
\[ PQ = \sqrt{(3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2} \]

### Step 3: Set the Distance Equal to \( \sqrt{5} \)
Since we know that the distance \( PQ \) is equal to \( \sqrt{5} \), we can set up the equation:
\[ \sqrt{(3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2} = \sqrt{5} \]
Squaring both sides gives:
\[ (3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2 = 5 \]

### Step 4: Expand and Simplify
Expanding each term:
\[ (3\lambda - 3)^2 = 9\lambda^2 - 18\lambda + 9 \]
\[ (2\lambda - 3)^2 = 4\lambda^2 - 12\lambda + 9 \]
\[ (2\lambda)^2 = 4\lambda^2 \]
Combining these, we get:
\[ 9\lambda^2 - 18\lambda + 9 + 4\lambda^2 - 12\lambda + 9 + 4\lambda^2 = 5 \]
This simplifies to:
\[ 17\lambda^2 - 30\lambda + 18 - 5 = 0 \]
\[ 17\lambda^2 - 30\lambda + 13 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \lambda = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 17 \cdot 13}}{2 \cdot 17} \]
Calculating the discriminant:
\[ 900 - 884 = 16 \]
Thus,
\[ \lambda = \frac{30 \pm 4}{34} \]
This gives us two values:
\[ \lambda_1 = \frac{34}{34} = 1, \quad \lambda_2 = \frac{26}{34} = \frac{13}{17} \]

### Step 6: Find Coordinates of \( Q \)
For \( \lambda = 1 \):
\[ Q(3(1) - 2, 2(1) - 1, 2(1) + 3) = Q(1, 1, 5) \]
For \( \lambda = \frac{13}{17} \):
\[ Q\left(3\left(\frac{13}{17}\right) - 2, 2\left(\frac{13}{17}\right) - 1, 2\left(\frac{13}{17}\right) + 3\right) = Q\left(\frac{39}{17} - \frac{34}{17}, \frac{26}{17} - \frac{17}{17}, \frac{26}{17} + \frac{51}{17}\right) = Q\left(\frac{5}{17}, \frac{9}{17}, \frac{77}{17}\right) \]

### Final Answer
The coordinates of the point \( Q \) that is at a distance of \( \sqrt{5} \) units from \( P(1, 2, 3) \) on the line are:
1. \( Q(1, 1, 5) \)
2. \( Q\left(\frac{5}{17}, \frac{9}{17}, \frac{77}{17}\right) \)