If the line `(x-1)/(2)=(y+1)/(3)=(z-1)/(4) and (x-3)/(1)=(y-k)/(2)=(z)/(1)` intersect, then k is equal to

Options

A. `-1`
B. `(2)/(9)`
C. `(9)/(2)`
D. `0`

Answer

To find the value of \( k \) for which the two lines intersect, we can follow these steps:

### Step 1: Write the equations of the lines in parametric form.

The first line \( L_1 \) is given by:
\[ \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} \]
Let \( t \) be the parameter. Then we can express the coordinates of points on this line as:
\[ x = 1 + 2t, \quad y = -1 + 3t, \quad z = 1 + 4t \]

The second line \( L_2 \) is given by:
\[ \frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1} \]
Let \( s \) be the parameter for this line. Then we can express the coordinates of points on this line as:
\[ x = 3 + s, \quad y = k + 2s, \quad z = s \]

### Step 2: Set the equations equal to each other.

Since the lines intersect, we can set the corresponding coordinates equal to each other:
1. \( 1 + 2t = 3 + s \)
2. \( -1 + 3t = k + 2s \)
3. \( 1 + 4t = s \)

### Step 3: Solve the equations.

From the third equation, we can express \( s \) in terms of \( t \):
\[ s = 1 + 4t \]

Now substitute \( s \) into the first equation:
\[ 1 + 2t = 3 + (1 + 4t) \]
This simplifies to:
\[ 1 + 2t = 4 + 4t \]
Rearranging gives:
\[ 2t - 4t = 4 - 1 \]
\[ -2t = 3 \implies t = -\frac{3}{2} \]

Now substitute \( t = -\frac{3}{2} \) back into the expression for \( s \):
\[ s = 1 + 4\left(-\frac{3}{2}\right) = 1 - 6 = -5 \]

### Step 4: Substitute \( t \) and \( s \) into the second equation.

Now substitute \( t = -\frac{3}{2} \) into the second equation:
\[ -1 + 3\left(-\frac{3}{2}\right) = k + 2(-5) \]
This simplifies to:
\[ -1 - \frac{9}{2} = k - 10 \]
Combining the left side gives:
\[ -\frac{2}{2} - \frac{9}{2} = k - 10 \]
\[ -\frac{11}{2} = k - 10 \]
Adding 10 to both sides:
\[ k = -\frac{11}{2} + 10 = -\frac{11}{2} + \frac{20}{2} = \frac{9}{2} \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{\frac{9}{2}} \]