The value of `k` such that `(x-4)/1=(y-2)/1=(z-k)/2` lies in the plane `2x-4y+z=7` is a. 7 b. -7 c. no real value d. 4

Options

A. 7
B. -7
C. no real value
D. 4

Explanation

To find the value of \( k \) such that the line given by the equation

\[ \frac{x-4}{1} = \frac{y-2}{1} = \frac{z-k}{2} \]

lies in the plane defined by

\[ 2x - 4y + z = 7, \]

we can follow these steps:

### Step 1: Parameterize the line

From the equation of the line, we can express \( x \), \( y \), and \( z \) in terms of a parameter \( t \):

\[ x = 4 + t, \]
\[ y = 2 + t, \]
\[ z = k + 2t. \]

### Step 2: Substitute into the plane equation

Now, we need to substitute \( x \), \( y \), and \( z \) into the plane equation \( 2x - 4y + z = 7 \):

\[ 2(4 + t) - 4(2 + t) + (k + 2t) = 7. \]

### Step 3: Simplify the equation

Expanding this gives:

\[ 8 + 2t - 8 - 4t + k + 2t = 7. \]

Combining like terms results in:

\[ k + 0t = 7. \]

This simplifies to:

\[ k = 7. \]

### Conclusion

Thus, the value of \( k \) such that the line lies in the plane is

\[ \boxed{7}. \]