Let `a=3hati + 2hatj + xhatk` and `b = hati - hatj + hatk`, for some real x. Then `|a+b| = r` is possible if.

Options

A. `0 lt r le sqrt(3)/2`
B. `sqrt(3)/2 lt r le 3sqrt(3/2)`
C. `3sqrt(3/2) lt r le 5sqrt(3/2)`
D. `r ge sqrt(17)`

Answer

To solve the problem, we need to find the magnitude of the vector sum \( a + b \) and determine the conditions under which this magnitude can be equal to \( r \).

### Step-by-Step Solution:

1. **Define the vectors**:
\[ a = 3\hat{i} + 2\hat{j} + x\hat{k} \]
\[ b = \hat{i} - \hat{j} + \hat{k} \]

2. **Calculate the vector sum \( a + b \)**:
\[ a + b = (3\hat{i} + 2\hat{j} + x\hat{k}) + (\hat{i} - \hat{j} + \hat{k}) \]
Combine the components:
\[ = (3 + 1)\hat{i} + (2 - 1)\hat{j} + (x + 1)\hat{k} \]
\[ = 4\hat{i} + 1\hat{j} + (x + 1)\hat{k} \]

3. **Find the magnitude of \( a + b \)**:
The magnitude \( |a + b| \) is given by:
\[ |a + b| = \sqrt{(4)^2 + (1)^2 + (x + 1)^2} \]
\[ = \sqrt{16 + 1 + (x + 1)^2} \]
\[ = \sqrt{17 + (x + 1)^2} \]

4. **Determine the conditions for \( r \)**:
Since \( x \) is a real number, \( (x + 1)^2 \) is always non-negative. Therefore:
\[ (x + 1)^2 \geq 0 \]
This implies:
\[ 17 + (x + 1)^2 \geq 17 \]
Taking the square root of both sides gives:
\[ \sqrt{17 + (x + 1)^2} \geq \sqrt{17} \]
Thus:
\[ |a + b| \geq \sqrt{17} \]
Therefore, \( r \) must satisfy:
\[ r \geq \sqrt{17} \]

### Conclusion:
The magnitude \( r \) is possible if \( r \geq \sqrt{17} \).