The position vector of the point in which the line joining the points `hati-2hatj+hatk` and `3hatk-2hatj` cuts the plane through the origin and the points `4hatj` and `2hati+hatk` is

Options

A. `5hati-10hatj+3hatk`
B. `1/5(6hati-10hatj+3hatk)`
C. `-6hati+10hatj-3hatk`
D. none of these

Explanation

To find the position vector of the point where the line joining the points \( \hat{i} - 2\hat{j} + \hat{k} \) and \( 3\hat{k} - 2\hat{j} \) intersects the plane defined by the origin, \( 4\hat{j} \), and \( 2\hat{i} + \hat{k} \), we can follow these steps:

### Step 1: Identify the Points
The points are:
- Point A: \( \hat{i} - 2\hat{j} + \hat{k} \) which corresponds to \( (1, -2, 1) \)
- Point B: \( 3\hat{k} - 2\hat{j} \) which corresponds to \( (0, -2, 3) \)
- Point C: \( 4\hat{j} \) which corresponds to \( (0, 4, 0) \)
- Point D: \( 2\hat{i} + \hat{k} \) which corresponds to \( (2, 0, 1) \)

### Step 2: Find the Equation of the Plane
The equation of the plane can be derived using the determinant method. The points involved are \( O(0, 0, 0) \), \( C(0, 4, 0) \), and \( D(2, 0, 1) \).

The equation of the plane can be expressed as:
\[ \begin{vmatrix} x & y & z & 1 \\ 0 & 4 & 0 & 1 \\ 2 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{vmatrix} = 0 \]

Calculating the determinant, we get:
\[ x(4 \cdot 1 - 0 \cdot 0) - y(0 \cdot 1 - 0 \cdot 0) + z(0 \cdot 2 - 0 \cdot 4) + 1(0 \cdot 0 - 0 \cdot 2) = 0 \]
This simplifies to:
\[ 4x + 8z = 0 \quad \text{or} \quad 4x + 8z = 0 \]

### Step 3: Parametrize the Line
The line joining points A and B can be parametrized as:
\[ \frac{x - 1}{0 - 1} = \frac{y + 2}{-2 + 2} = \frac{z - 1}{3 - 1} \]
This simplifies to:
\[ x = 1 + t, \quad y = -2, \quad z = 1 + 2t \]

### Step 4: Substitute into the Plane Equation
Substituting \( x \) and \( z \) into the plane equation \( 4x + 8z = 0 \):
\[ 4(1 + t) + 8(1 + 2t) = 0 \]
Expanding this gives:
\[ 4 + 4t + 8 + 16t = 0 \]
Combining like terms:
\[ 20t + 12 = 0 \]
Solving for \( t \):
\[ t = -\frac{12}{20} = -\frac{3}{5} \]

### Step 5: Find the Coordinates of the Intersection Point
Now substituting \( t = -\frac{3}{5} \) back into the parametric equations:
\[ x = 1 - \frac{3}{5} = \frac{2}{5} \]
\[ y = -2 \]
\[ z = 1 + 2(-\frac{3}{5}) = 1 - \frac{6}{5} = -\frac{1}{5} \]

### Step 6: Write the Position Vector
Thus, the position vector of the point of intersection is:
\[ \frac{2}{5}\hat{i} - 2\hat{j} - \frac{1}{5}\hat{k} \]

### Final Answer
The position vector is:
\[ \frac{2}{5}\hat{i} - 2\hat{j} - \frac{1}{5}\hat{k} \]