`vecA=2hati+hatj`,`B=3hatj-hatk` and `vecC=6hati-2hatk`.
value of `vecA-2vecB+3vecC` would be

Options

A. `20hati+5hatj+4hatk`
B. `20hati-5hatj-4hatk`
C. `4hati+5hatj+20hatk`
D. `5hati+4hatj+10hatk`

Explanation

To solve the problem, we need to compute the expression \( \vec{A} - 2\vec{B} + 3\vec{C} \) using the given vectors:

1. **Given Vectors:**
- \( \vec{A} = 2\hat{i} + \hat{j} \)
- \( \vec{B} = 3\hat{j} - \hat{k} \)
- \( \vec{C} = 6\hat{i} - 2\hat{k} \)

2. **Substituting the Vectors:**
We substitute the vectors into the expression:
\[ \vec{A} - 2\vec{B} + 3\vec{C} = (2\hat{i} + \hat{j}) - 2(3\hat{j} - \hat{k}) + 3(6\hat{i} - 2\hat{k}) \]

3. **Calculating \( -2\vec{B} \):**
\[ -2\vec{B} = -2(3\hat{j} - \hat{k}) = -6\hat{j} + 2\hat{k} \]

4. **Calculating \( 3\vec{C} \):**
\[ 3\vec{C} = 3(6\hat{i} - 2\hat{k}) = 18\hat{i} - 6\hat{k} \]

5. **Combining All Parts:**
Now we combine all parts:
\[ \vec{A} - 2\vec{B} + 3\vec{C} = (2\hat{i} + \hat{j}) + (-6\hat{j} + 2\hat{k}) + (18\hat{i} - 6\hat{k}) \]

6. **Grouping Like Terms:**
- For \( \hat{i} \): \( 2\hat{i} + 18\hat{i} = 20\hat{i} \)
- For \( \hat{j} \): \( \hat{j} - 6\hat{j} = -5\hat{j} \)
- For \( \hat{k} \): \( 2\hat{k} - 6\hat{k} = -4\hat{k} \)

7. **Final Result:**
Combining these results, we get:
\[ \vec{A} - 2\vec{B} + 3\vec{C} = 20\hat{i} - 5\hat{j} - 4\hat{k} \]

Thus, the final answer is:
\[ \vec{A} - 2\vec{B} + 3\vec{C} = 20\hat{i} - 5\hat{j} - 4\hat{k} \]