If the line `(x+1)/3=(y-2)/4=(z+6)/5` is parallel to the planes `2x-3y+kz=0`, then the value of k is

Options

A. `5/6`
B. `6/5`
C. `3/4`
D. `4/5`

Explanation

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

\[ \frac{x+1}{3} = \frac{y-2}{4} = \frac{z+6}{5} \]

is parallel to the plane defined by

\[ 2x - 3y + kz = 0. \]

### Step 1: Identify the direction ratios of the line
From the equation of the line, we can extract the direction ratios. The line can be represented in vector form as:

\[ \mathbf{r} = (-1, 2, -6) + t(3, 4, 5) \]

where \( t \) is a parameter. Thus, the direction ratios of the line are:

\[ \mathbf{b} = (3, 4, 5). \]

### Step 2: Identify the normal vector of the plane
The equation of the plane is given as:

\[ 2x - 3y + kz = 0. \]

From this equation, we can identify the normal vector \( \mathbf{n} \) of the plane as:

\[ \mathbf{n} = (2, -3, k). \]

### Step 3: Use the 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 the dot product of the direction vector \( \mathbf{b} \) and the normal vector \( \mathbf{n} \) must be zero:

\[ \mathbf{n} \cdot \mathbf{b} = 0. \]

### Step 4: Calculate the dot product
Calculating the dot product:

\[ \mathbf{n} \cdot \mathbf{b} = (2, -3, k) \cdot (3, 4, 5) = 2 \cdot 3 + (-3) \cdot 4 + k \cdot 5. \]

This simplifies to:

\[ 6 - 12 + 5k = 0. \]

### Step 5: Solve for \( k \)
Now, we simplify the equation:

\[ 6 - 12 + 5k = 0 \implies -6 + 5k = 0 \implies 5k = 6 \implies k = \frac{6}{5}. \]

Thus, the value of \( k \) is

\[ \boxed{\frac{6}{5}}. \]