The vector `vecB=5hati+2hatj-Shatk` is perpendicular to the vector `vecA=3hati+hatj+2hatk` if S=

Explanation

To solve the problem, we need to find the value of \( S \) such that the vector \( \vec{B} = 5\hat{i} + 2\hat{j} - S\hat{k} \) is perpendicular to the vector \( \vec{A} = 3\hat{i} + \hat{j} + 2\hat{k} \).

### Step-by-Step Solution:

1. **Understand the Condition for Perpendicularity:**
Two vectors are perpendicular if their dot product is equal to zero. Therefore, we need to compute the dot product \( \vec{A} \cdot \vec{B} \) and set it equal to zero.

2. **Write the Dot Product:**
The dot product \( \vec{A} \cdot \vec{B} \) is calculated as follows:
\[ \vec{A} \cdot \vec{B} = (3\hat{i} + \hat{j} + 2\hat{k}) \cdot (5\hat{i} + 2\hat{j} - S\hat{k}) \]

3. **Calculate the Dot Product:**
Using the distributive property of the dot product:
\[ \vec{A} \cdot \vec{B} = 3 \cdot 5 + 1 \cdot 2 + 2 \cdot (-S) \]
\[ = 15 + 2 - 2S \]

4. **Set the Dot Product Equal to Zero:**
Since \( \vec{A} \) and \( \vec{B} \) are perpendicular, we set the dot product equal to zero:
\[ 15 + 2 - 2S = 0 \]

5. **Simplify the Equation:**
Combine the constants:
\[ 17 - 2S = 0 \]

6. **Solve for \( S \):**
Rearranging the equation gives:
\[ 2S = 17 \]
\[ S = \frac{17}{2} \]

### Final Answer:
The value of \( S \) is \( \frac{17}{2} \).