From a point A with positon vector `p(hati+hatj+hatk),AB` and `AC` are drawn perpendicular to the lines `vecr=hatk+lamda(hati+hatj)` an `vecr=-hatk+mu(hati-hatj)`, respectively. A value of `p` is equal to

Options

A. `-2`
B. `-1`
C. `sqrt(2)`
D. all of these

Explanation

To solve the problem, we need to find the value of \( p \) given the conditions about the lines and the perpendiculars drawn from point \( A \).

### Step-by-Step Solution:

1. **Identify the Position Vector of Point A**:
The position vector of point \( A \) is given as:
\[ \vec{A} = p \hat{i} + p \hat{j} + p \hat{k} \]
This can be expressed as:
\[ \vec{A} = (p, p, p) \]

2. **Write the Equations of the Lines**:
The equations of the lines are given as:
- Line 1:
\[ \vec{r} = \hat{k} + \lambda (\hat{i} + \hat{j}) \]
This can be expanded to:
\[ \vec{r_1} = (0 + \lambda, 0 + \lambda, 1) = (\lambda, \lambda, 1) \]
- Line 2:
\[ \vec{r} = -\hat{k} + \mu (\hat{i} - \hat{j}) \]
This can be expanded to:
\[ \vec{r_2} = (\mu, -\mu, -1) \]

3. **Find the Position Vectors of Points B and C**:
- The position vector of point \( B \) on line 1 is:
\[ \vec{B} = (\lambda, \lambda, 1) \]
- The position vector of point \( C \) on line 2 is:
\[ \vec{C} = (\mu, -\mu, -1) \]

4. **Determine the Vector AB**:
The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (\lambda - p, \lambda - p, 1 - p) \]

5. **Determine the Direction Ratios of Line 1**:
The direction ratios of line 1 are \( (1, 1, 0) \) since it is parallel to \( \hat{i} + \hat{j} \).

6. **Set Up the Perpendicular Condition for AB and Line 1**:
For \( \vec{AB} \) to be perpendicular to line 1, their dot product must equal zero:
\[ (\lambda - p, \lambda - p, 1 - p) \cdot (1, 1, 0) = 0 \]
This simplifies to:
\[ (\lambda - p) + (\lambda - p) = 0 \implies 2(\lambda - p) = 0 \implies \lambda = p \]

7. **Determine the Vector AC**:
The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} = (\mu - p, -\mu - p, -1 - p) \]

8. **Determine the Direction Ratios of Line 2**:
The direction ratios of line 2 are \( (1, -1, 0) \).

9. **Set Up the Perpendicular Condition for AC and Line 2**:
For \( \vec{AC} \) to be perpendicular to line 2, their dot product must equal zero:
\[ (\mu - p, -\mu - p, -1 - p) \cdot (1, -1, 0) = 0 \]
This simplifies to:
\[ (\mu - p) - (-\mu - p) = 0 \implies \mu - p + \mu + p = 0 \implies 2\mu = 0 \implies \mu = 0 \]

10. **Conclusion**:
From the conditions derived, we have:
- \( \lambda = p \)
- \( \mu = 0 \)

Since \( p \) can take any scalar value, the final answer is that \( p \) can be any real number.

### Final Answer:
The value of \( p \) can be any scalar value.