If three vectors `V_(1)=alphai+j+k, V_(2)=i+betaj-2k" and "V_(3)=i+j` are coplanar, and `V_(1)" and "V_(3)` are perpendicular, then the vector `V_(1) times V_(2)` is:

Options

A. `-i+j+2k`
B. `i-j+2k`
C. `-i+j`
D. `2i-2j+k`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the vectors
We are given three vectors:
- \( V_1 = \alpha i + j + k \)
- \( V_2 = i + \beta j - 2k \)
- \( V_3 = i + j \)

### Step 2: Use the coplanarity condition
The vectors \( V_1, V_2, V_3 \) are coplanar if their scalar triple product is zero:
\[ V_1 \cdot (V_2 \times V_3) = 0 \]
This can be expressed as the determinant:
\[ \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \beta & -2 \\ 1 & 1 & 0 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ = \alpha \begin{vmatrix} \beta & -2 \\ 1 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} + 1 \begin{vmatrix} 1 & \beta \\ 1 & 1 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} \beta & -2 \\ 1 & 0 \end{vmatrix} = 0 - (-2) = 2 \)
2. \( \begin{vmatrix} 1 & -2 \\ 1 & 0 \end{vmatrix} = 0 - (-2) = 2 \)
3. \( \begin{vmatrix} 1 & \beta \\ 1 & 1 \end{vmatrix} = 1 - \beta \)

Putting it all together:
\[ \alpha \cdot 2 - 1 \cdot 2 + (1 - \beta) = 0 \]
This simplifies to:
\[ 2\alpha - 2 + 1 - \beta = 0 \quad \text{(1)} \]

### Step 4: Use the perpendicularity condition
Vectors \( V_1 \) and \( V_3 \) are perpendicular, so:
\[ V_1 \cdot V_3 = 0 \]
Calculating the dot product:
\[ (\alpha i + j + k) \cdot (i + j) = \alpha \cdot 1 + 1 \cdot 1 + 0 = \alpha + 1 = 0 \]
Thus, we have:
\[ \alpha + 1 = 0 \quad \Rightarrow \quad \alpha = -1 \quad \text{(2)} \]

### Step 5: Substitute \(\alpha\) into equation (1)
Substituting \(\alpha = -1\) into equation (1):
\[ 2(-1) - 2 + 1 - \beta = 0 \]
This simplifies to:
\[ -2 - 2 + 1 - \beta = 0 \quad \Rightarrow \quad -3 - \beta = 0 \quad \Rightarrow \quad \beta = -3 \]

### Step 6: Write the vectors with found values
Now substituting \(\alpha\) and \(\beta\) back into the vectors:
- \( V_1 = -i + j + k \)
- \( V_2 = i - 3j - 2k \)
- \( V_3 = i + j \)

### Step 7: Calculate \( V_1 \times V_2 \)
Now we compute the cross product \( V_1 \times V_2 \):
\[ \begin{vmatrix} i & j & k \\ -1 & 1 & 1 \\ 1 & -3 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ = i \begin{vmatrix} 1 & 1 \\ -3 & -2 \end{vmatrix} - j \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} + k \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ -3 & -2 \end{vmatrix} = (1)(-2) - (1)(-3) = -2 + 3 = 1 \)
2. \( \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (1)(1) = 2 - 1 = 1 \)
3. \( \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (1)(1) = 3 - 1 = 2 \)

Putting it all together:
\[ V_1 \times V_2 = i(1) - j(1) + k(2) = i - j + 2k \]

### Final Answer
Thus, the vector \( V_1 \times V_2 \) is:
\[ \boxed{i - j + 2k} \]