The line `(x-1)/(3)=(y-3)/(k)=(z+1)/(5)` is perpendicular to the plane 6x-8y+10z-7=0 if k =

Options

A. 4
B. `-4`
C. 3
D. 2

Answer

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

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

is perpendicular to the plane defined by the equation

\[ 6x - 8y + 10z - 7 = 0. \]

### Step-by-Step Solution:

1. **Identify the Direction Vector of the Line:**
The line can be expressed in parametric form. The direction ratios of the line are given by the coefficients of \( x, y, z \) in the equation:
\[ \text{Direction vector of the line} = \vec{b} = \langle 3, k, 5 \rangle. \]

2. **Identify the Normal Vector of the Plane:**
The normal vector of the plane can be derived from the coefficients of \( x, y, z \) in the plane equation:
\[ \text{Normal vector of the plane} = \vec{n} = \langle 6, -8, 10 \rangle. \]

3. **Set Up the Condition for Perpendicularity:**
For the line to be perpendicular to the plane, the direction vector of the line must be parallel to the normal vector of the plane. This can be expressed as:
\[ \vec{n} = \lambda \vec{b} \]
for some scalar \( \lambda \).

4. **Write the Vector Equation:**
This gives us the following equations by comparing the components:
\[ 6 = 3\lambda, \quad -8 = k\lambda, \quad 10 = 5\lambda. \]

5. **Solve for \( \lambda \):**
From the first equation:
\[ \lambda = \frac{6}{3} = 2. \]
From the third equation:
\[ \lambda = \frac{10}{5} = 2. \]
Both equations confirm that \( \lambda = 2 \).

6. **Substitute \( \lambda \) to Find \( k \):**
Substitute \( \lambda \) into the second equation:
\[ -8 = k \cdot 2 \implies k = \frac{-8}{2} = -4. \]

### Final Answer:
Thus, the value of \( k \) is

\[ \boxed{-4}. \]