If the four points,whose position vectors are `3hati-4hat j+2hat k,hat i+2hat j-hat k,-2hat i-hat j+3hat k` and `5hat i-2alphahat j+4hat k` are coplanar,then `alpha` is equal to

Options

A. `(107)/(17)`
B. `(73)/(17)`
C. `-(73)/(17)`
D. `-(107)/(17)`

Answer

To determine the value of \( \alpha \) such that the four points with given position vectors are coplanar, we can use the concept of determinants. The position vectors of the points are:

1. \( \mathbf{A} = 3\hat{i} - 4\hat{j} + 2\hat{k} \)
2. \( \mathbf{B} = \hat{i} + 2\hat{j} - \hat{k} \)
3. \( \mathbf{C} = -2\hat{i} - \hat{j} + 3\hat{k} \)
4. \( \mathbf{D} = 5\hat{i} - 2\alpha\hat{j} + 4\hat{k} \)

The points are coplanar if the scalar triple product of the vectors formed by these points is zero. We can express this condition using a determinant:

\[ \begin{vmatrix} 3 & -4 & 2 \\ 1 & 2 & -1 \\ -2 & -1 & 3 \\ 5 & -2\alpha & 4 \end{vmatrix} = 0 \]

### Step 1: Set up the determinant
We will set up the determinant using the coordinates of the vectors:

\[ \begin{vmatrix} 3 & -4 & 2 \\ 1 & 2 & -1 \\ -2 & -1 & 3 \\ 5 & -2\alpha & 4 \end{vmatrix} \]

### Step 2: Expand the determinant
We can expand the determinant using the first row:

\[ = 3 \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} - (-4) \begin{vmatrix} 1 & -1 \\ -2 & 3 \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} - 5 \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} = (2)(3) - (-1)(-1) = 6 - 1 = 5 \)

2. \( \begin{vmatrix} 1 & -1 \\ -2 & 3 \end{vmatrix} = (1)(3) - (-1)(-2) = 3 - 2 = 1 \)

3. \( \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} = (1)(-1) - (2)(-2) = -1 + 4 = 3 \)

### Step 4: Substitute back into the determinant
Now substituting back into the determinant:

\[ = 3(5) + 4(1) + 2(3) - 5(3) \]

Calculating this gives:

\[ = 15 + 4 + 6 - 15 = 10 - 15 + 4 = 4 - 15 = -11 + 4 = -7 \]

### Step 5: Include \( \alpha \) and set the determinant to zero
Now we include \( \alpha \):

\[ = 3(5) + 4(1) + 2(3) - 5(3) + 2\alpha(-6) = 0 \]

This simplifies to:

\[ -6\alpha + 10 = 0 \]

### Step 6: Solve for \( \alpha \)
Solving for \( \alpha \):

\[ -6\alpha = -10 \implies \alpha = \frac{10}{6} = \frac{5}{3} \]

Thus, the value of \( \alpha \) is:

\[ \alpha = \frac{5}{3} \]

### Final Answer
The value of \( \alpha \) is \( \frac{5}{3} \).