If the line `(x-1)/2=(y+3)/1=(z-5)/(-1)` is parallel to the
plane `px + 3y - z + 5 = 0 `, then the value of p -

Options

A. 2
B. `-2`
C. `1/2`
D. `1/3`

Answer

To solve the problem, we need to find the value of \( p \) such that the line given by the equation

\[ \frac{x-1}{2} = \frac{y+3}{1} = \frac{z-5}{-1} \]

is parallel to the plane described by the equation

\[ px + 3y - z + 5 = 0. \]

### Step-by-Step Solution:

1. **Identify the Direction Ratios of the Line**:
The line can be expressed in parametric form. From the equation, we can identify the direction ratios (or direction vector) of the line as:
\[ \mathbf{b} = (2, 1, -1). \]

2. **Identify the Normal Vector of the Plane**:
The normal vector to the plane given by the equation \( px + 3y - z + 5 = 0 \) can be derived from the coefficients of \( x \), \( y \), and \( z \). Thus, the normal vector is:
\[ \mathbf{n} = (p, 3, -1). \]

3. **Condition for Parallelism**:
For the line to be parallel to the plane, the direction vector of the line must be perpendicular to the normal vector of the plane. This means that their dot product must equal zero:
\[ \mathbf{b} \cdot \mathbf{n} = 0. \]

4. **Calculate the Dot Product**:
We compute the dot product:
\[ \mathbf{b} \cdot \mathbf{n} = (2, 1, -1) \cdot (p, 3, -1) = 2p + 1 \cdot 3 + (-1)(-1). \]
Simplifying this gives:
\[ 2p + 3 + 1 = 0. \]
Therefore:
\[ 2p + 4 = 0. \]

5. **Solve for \( p \)**:
Rearranging the equation gives:
\[ 2p = -4 \implies p = -2. \]

### Final Answer:
The value of \( p \) is \( -2 \).

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