If vectors `vecA=2hati+3hatj+4hatk, vecB=hati+hatj+5hatk and vecC` form a left handed system then `vecC` is (A) `11hati-6hatj-hatk` (B) `-11hati+6hatj+hatk` (C) `-11hati+6hatj-hatk` (D) `-11hati+6hatj-hatk`

Options

A. `11 hati - 6 hatj - hatk`
B. `-11 hati - 6 hatj - hatk`
C. `-11 hati - 6 hatj + hatk`
D. `-11 hati + 6 hatj - hatk`

Explanation

To solve the problem, we need to find the vector \( \vec{C} \) such that the vectors \( \vec{A}, \vec{B}, \) and \( \vec{C} \) form a left-handed system. This means that the scalar triple product \( \vec{A} \cdot (\vec{B} \times \vec{C}) \) must be negative.

### Step-by-step Solution:

1. **Define the vectors**:
\[ \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
\[ \vec{B} = \hat{i} + \hat{j} + 5\hat{k} \]
Let \( \vec{C} = x\hat{i} + y\hat{j} + z\hat{k} \).

2. **Calculate the cross product \( \vec{A} \times \vec{B} \)**:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 1 & 5 \end{vmatrix} \]

Expanding this determinant:
\[ = \hat{i} \left( 3 \cdot 5 - 4 \cdot 1 \right) - \hat{j} \left( 2 \cdot 5 - 4 \cdot 1 \right) + \hat{k} \left( 2 \cdot 1 - 3 \cdot 1 \right) \]
\[ = \hat{i} (15 - 4) - \hat{j} (10 - 4) + \hat{k} (2 - 3) \]
\[ = 11\hat{i} - 6\hat{j} - 1\hat{k} \]

Thus,
\[ \vec{A} \times \vec{B} = 11\hat{i} - 6\hat{j} - \hat{k} \]

3. **Dot product with \( \vec{C} \)**:
We need to find \( \vec{C} \) such that:
\[ \vec{A} \times \vec{B} \cdot \vec{C} < 0 \]
This gives:
\[ (11\hat{i} - 6\hat{j} - \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) < 0 \]
\[ = 11x - 6y - z < 0 \]

4. **Check the options**:
We need to evaluate the expression \( 11x - 6y - z \) for each option to see which one satisfies the inequality.

- **Option A**: \( \vec{C} = 11\hat{i} - 6\hat{j} - \hat{k} \)
\[ x = 11, y = -6, z = -1 \Rightarrow 11(11) - 6(-6) - (-1) = 121 + 36 + 1 = 158 \quad (\text{not valid}) \]

- **Option B**: \( \vec{C} = -11\hat{i} + 6\hat{j} + \hat{k} \)
\[ x = -11, y = 6, z = 1 \Rightarrow 11(-11) - 6(6) - 1 = -121 - 36 - 1 = -158 \quad (\text{valid}) \]

- **Option C**: \( \vec{C} = -11\hat{i} + 6\hat{j} - \hat{k} \)
\[ x = -11, y = 6, z = -1 \Rightarrow 11(-11) - 6(6) - (-1) = -121 - 36 + 1 = -156 \quad (\text{valid}) \]

- **Option D**: \( \vec{C} = -11\hat{i} + 6\hat{j} - \hat{k} \)
\[ x = -11, y = 6, z = -1 \Rightarrow 11(-11) - 6(6) - (-1) = -121 - 36 + 1 = -156 \quad (\text{valid}) \]

5. **Conclusion**:
The correct answer is **Option B**: \( \vec{C} = -11\hat{i} + 6\hat{j} + \hat{k} \).