If `vecA,vecB,vecC` are three vectors respectively given by `2hati+hatk, hati+hatj+hatk and 4hati-3hatj+7hatk, ` then the vector `vecR` which satisfies the relations `vecRxxvecB=vecCxxvecB and vecR.vecA=0` is (A) `2hati-8hatj+2hatk` (B) `hati-4hatj+2hatk` (C) `-hati-8hatj+2hatk` (D) none of these

Answer

To solve the problem, we need to find the vector \(\vec{R}\) that satisfies the conditions \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\) and \(\vec{R} \cdot \vec{A} = 0\). Let's break this down step by step.

### Step 1: Define the vectors
Given:
- \(\vec{A} = 2\hat{i} + \hat{k}\)
- \(\vec{B} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{C} = 4\hat{i} - 3\hat{j} + 7\hat{k}\)

### Step 2: Find \(\vec{R}\) in terms of components
Let \(\vec{R} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 3: Use the condition \(\vec{R} \cdot \vec{A} = 0\)
Calculate the dot product:
\[ \vec{R} \cdot \vec{A} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 2x + z = 0 \]
This gives us our first equation:
\[ z = -2x \quad \text{(1)} \]

### Step 4: Find \(\vec{C} \times \vec{B}\)
Now we need to calculate \(\vec{C} \times \vec{B}\):
\[ \vec{C} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -3 & 7 \\ 1 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-3)(1) - (7)(1)) - \hat{j}((4)(1) - (7)(1)) + \hat{k}((4)(1) - (-3)(1)) \]
\[ = \hat{i}(-3 - 7) - \hat{j}(4 - 7) + \hat{k}(4 + 3) \]
\[ = -10\hat{i} + 3\hat{j} + 7\hat{k} \]

### Step 5: Use the condition \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\)
Now we need to compute \(\vec{R} \times \vec{B}\):
\[ \vec{R} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(y(1) - z(1)) - \hat{j}(x(1) - z(1)) + \hat{k}(x(1) - y(1)) \]
\[ = (y - z)\hat{i} - (x - z)\hat{j} + (x - y)\hat{k} \]

### Step 6: Set the components equal to \(\vec{C} \times \vec{B}\)
Equating the components:
1. \(y - z = -10\) \quad (2)
2. \(- (x - z) = 3 \implies x - z = -3 \quad (3)\)
3. \(x - y = 7 \quad (4)\)

### Step 7: Substitute \(z\) from equation (1) into equations (2), (3), and (4)
Substituting \(z = -2x\) into equations (2), (3), and (4):
1. \(y - (-2x) = -10 \implies y + 2x = -10 \quad (5)\)
2. \(x - (-2x) = -3 \implies 3x = -3 \implies x = -1 \quad (6)\)
3. \(x - y = 7 \quad (4)\)

### Step 8: Solve for \(y\) and \(z\)
From equation (6):
\[ x = -1 \]
Substituting \(x = -1\) into equation (1):
\[ z = -2(-1) = 2 \]
Substituting \(x = -1\) into equation (4):
\[ -1 - y = 7 \implies y = -8 \]

### Step 9: Write the final vector \(\vec{R}\)
Thus, we have:
\[ \vec{R} = -1\hat{i} - 8\hat{j} + 2\hat{k} \]

### Conclusion
The vector \(\vec{R}\) is:
\[ \vec{R} = -\hat{i} - 8\hat{j} + 2\hat{k} \]
This corresponds to option (C).