The lines
`vecr=(2hati-3hatj+7hatk)+lamda(2hati+phatj+5hatk)`
and `vecr=(hati+2hatj+3hatk)+mu(3hati+phatj+phatk)` are perpendicular it `p=`

Options

A. `-2`
B. 2
C. 5
D. `-3`

Answer

To solve the problem, we need to determine the value of \( p \) such that the two given lines are perpendicular.

### Step-by-step Solution:

1. **Identify the Direction Vectors:**
The first line is given by:
\[ \vec{r} = (2\hat{i} - 3\hat{j} + 7\hat{k}) + \lambda(2\hat{i} + p\hat{j} + 5\hat{k}) \]
The direction vector \( \vec{b_1} \) of the first line is:
\[ \vec{b_1} = 2\hat{i} + p\hat{j} + 5\hat{k} \]

The second line is given by:
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(3\hat{i} + p\hat{j} + p\hat{k}) \]
The direction vector \( \vec{b_2} \) of the second line is:
\[ \vec{b_2} = 3\hat{i} + p\hat{j} + p\hat{k} \]

2. **Condition for Perpendicular Lines:**
For the lines to be perpendicular, the dot product of their direction vectors must be zero:
\[ \vec{b_1} \cdot \vec{b_2} = 0 \]

3. **Calculate the Dot Product:**
We compute the dot product:
\[ \vec{b_1} \cdot \vec{b_2} = (2\hat{i} + p\hat{j} + 5\hat{k}) \cdot (3\hat{i} + p\hat{j} + p\hat{k}) \]
Expanding this, we get:
\[ = 2 \cdot 3 + p \cdot p + 5 \cdot p \]
\[ = 6 + p^2 + 5p \]

4. **Set the Dot Product to Zero:**
We set the equation to zero:
\[ 6 + p^2 + 5p = 0 \]
Rearranging gives us:
\[ p^2 + 5p + 6 = 0 \]

5. **Factor the Quadratic Equation:**
We can factor the quadratic:
\[ (p + 2)(p + 3) = 0 \]

6. **Solve for \( p \):**
Setting each factor to zero gives:
\[ p + 2 = 0 \quad \Rightarrow \quad p = -2 \]
\[ p + 3 = 0 \quad \Rightarrow \quad p = -3 \]

### Final Answer:
Thus, the values of \( p \) that make the lines perpendicular are:
\[ p = -2 \quad \text{or} \quad p = -3 \]