A point at a distance of `sqrt(6)` from the origin which
lies on the straight line `(x-1)/1 = (y-2)/2=(z+1)/3` will be

Options

A. (1, -1, 2)
B. (1, 2, -1)
C. `(5/7, 10/7, -6/7)`
D. `(5/7, 2/7, -6/7)`

Answer

To solve the problem, we need to find a point that is at a distance of \(\sqrt{6}\) from the origin and lies on the line defined by the equations \(\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+1}{3}\).

### Step-by-Step Solution:

1. **Define the point on the line:**
Let the parameter \(\lambda\) be equal to the common ratio in the line equation. Then, we can express the coordinates of any point on the line as:
\[ x = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = -1 + 3\lambda \]

2. **Distance from the origin:**
The distance \(d\) from the origin to the point \((x, y, z)\) is given by the formula:
\[ d = \sqrt{x^2 + y^2 + z^2} \]
We want this distance to be \(\sqrt{6}\). Therefore, we set up the equation:
\[ \sqrt{(1 + \lambda)^2 + (2 + 2\lambda)^2 + (-1 + 3\lambda)^2} = \sqrt{6} \]

3. **Square both sides:**
Squaring both sides to eliminate the square root gives:
\[ (1 + \lambda)^2 + (2 + 2\lambda)^2 + (-1 + 3\lambda)^2 = 6 \]

4. **Expand the squares:**
Now, we expand each term:
\[ (1 + \lambda)^2 = 1 + 2\lambda + \lambda^2 \]
\[ (2 + 2\lambda)^2 = 4 + 8\lambda + 4\lambda^2 \]
\[ (-1 + 3\lambda)^2 = 1 - 6\lambda + 9\lambda^2 \]

5. **Combine the terms:**
Adding these together, we have:
\[ 1 + 2\lambda + \lambda^2 + 4 + 8\lambda + 4\lambda^2 + 1 - 6\lambda + 9\lambda^2 = 6 \]
Simplifying this gives:
\[ (1 + 4 + 1) + (2\lambda + 8\lambda - 6\lambda) + (1 + 4 + 9)\lambda^2 = 6 \]
\[ 6 + 4\lambda + 14\lambda^2 = 6 \]

6. **Set the equation to zero:**
Rearranging gives:
\[ 14\lambda^2 + 4\lambda + 6 - 6 = 0 \]
\[ 14\lambda^2 + 4\lambda = 0 \]

7. **Factor out common terms:**
Factoring out \(2\lambda\):
\[ 2\lambda(7\lambda + 2) = 0 \]

8. **Solve for \(\lambda\):**
This gives us two solutions:
\[ \lambda = 0 \quad \text{or} \quad 7\lambda + 2 = 0 \implies \lambda = -\frac{2}{7} \]

9. **Find the points:**
Substitute \(\lambda = 0\):
\[ P_1 = (1 + 0, 2 + 2(0), -1 + 3(0)) = (1, 2, -1) \]
Substitute \(\lambda = -\frac{2}{7}\):
\[ P_2 = \left(1 - \frac{2}{7}, 2 - \frac{4}{7}, -1 - \frac{6}{7}\right) = \left(\frac{5}{7}, \frac{10}{7}, -\frac{13}{7}\right) \]

### Final Points:
The two points that are at a distance of \(\sqrt{6}\) from the origin and lie on the given line are:
1. \(P_1 = (1, 2, -1)\)
2. \(P_2 = \left(\frac{5}{7}, \frac{10}{7}, -\frac{13}{7}\right)\)