If the line `(x-4)/(1)=(y-2)/(1)=(z-k)/(2)` lies exactly on the plane `2x-4y+z=7`, the value of k is

Answer

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

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

lies exactly on the plane defined by

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

we can follow these steps:

### Step 1: Parametrize the line
From the given 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 the parametric equations into the plane equation
Next, we 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
Now, we simplify the left-hand side:

\[ 2(4) + 2t - 4(2) - 4t + k + 2t = 7, \]
\[ 8 + 2t - 8 - 4t + k + 2t = 7, \]
\[ k + 0t = 7. \]

This simplifies to:

\[ k = 7. \]

### Conclusion
Thus, the value of \( k \) is

\[ \boxed{7}. \]