The position vector of a point at a distance of `3sqrt(11)` units from `hati-hatj+2hatk` on a line passing through the points `hati-hatj+2hatk` and parallel to the vector `3hati+hatj+hatk` is

Options

A. `10hati+2hatj-5hatk`
B. `-8hati-4hatj-hatk`
C. `8hati+4hatj+hatk`
D. `-10hati-2hatj-5hatk`

Explanation

To solve the problem, we need to find the position vector of a point that is at a distance of \(3\sqrt{11}\) units from the point represented by the vector \(\hat{i} - \hat{j} + 2\hat{k}\) on a line that passes through this point and is parallel to the vector \(3\hat{i} + \hat{j} + \hat{k}\).

### Step-by-step Solution:

1. **Identify the Given Point and Direction Vector**:
- The given point \(P\) has the position vector \(\mathbf{a} = \hat{i} - \hat{j} + 2\hat{k}\).
- The direction vector of the line is \(\mathbf{b} = 3\hat{i} + \hat{j} + \hat{k}\).

2. **Equation of the Line**:
- The parametric equations of the line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
- This gives us:
\[ \mathbf{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(3\hat{i} + \hat{j} + \hat{k}) \]
- Simplifying this, we have:
\[ \mathbf{r} = (1 + 3\lambda)\hat{i} + (-1 + \lambda)\hat{j} + (2 + \lambda)\hat{k} \]

3. **Coordinates of Point Q**:
- Let the coordinates of point \(Q\) be:
\[ Q(1 + 3\lambda, -1 + \lambda, 2 + \lambda) \]

4. **Distance Formula**:
- The distance \(d\) between points \(P\) and \(Q\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
- Substituting the coordinates of \(P\) and \(Q\):
\[ d = \sqrt{((1 + 3\lambda) - 1)^2 + ((-1 + \lambda) + 1)^2 + ((2 + \lambda) - 2)^2} \]
- This simplifies to:
\[ d = \sqrt{(3\lambda)^2 + \lambda^2 + \lambda^2} = \sqrt{9\lambda^2 + 2\lambda^2} = \sqrt{11\lambda^2} \]

5. **Set the Distance Equal to \(3\sqrt{11}\)**:
- We know that this distance is equal to \(3\sqrt{11}\):
\[ \sqrt{11\lambda^2} = 3\sqrt{11} \]
- Squaring both sides gives:
\[ 11\lambda^2 = 9 \cdot 11 \]
- Dividing by \(11\):
\[ \lambda^2 = 9 \]
- Therefore, \(\lambda = 3\) or \(\lambda = -3\).

6. **Finding Coordinates for Both Values of \(\lambda\)**:
- For \(\lambda = 3\):
\[ Q(1 + 3 \cdot 3, -1 + 3, 2 + 3) = Q(10, 2, 5) \]
- For \(\lambda = -3\):
\[ Q(1 + 3 \cdot (-3), -1 - 3, 2 - 3) = Q(-8, -4, -1) \]

7. **Position Vectors**:
- The position vectors for the points \(Q\) are:
- For \(\lambda = 3\): \(10\hat{i} + 2\hat{j} + 5\hat{k}\)
- For \(\lambda = -3\): \(-8\hat{i} - 4\hat{j} - \hat{k}\)

### Final Answer:
The position vectors of the points are:
1. \(10\hat{i} + 2\hat{j} + 5\hat{k}\)
2. \(-8\hat{i} - 4\hat{j} - \hat{k}\)