To solve the problem, we need to find the value of \( k \) such that the two given lines intersect, and then we will find the equation of the plane containing these lines.
### Step 1: Identify the lines
The two lines are given in symmetric form:
1. Line 1: \( \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \)
2. Line 2: \( \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} \)
### Step 2: Write the parametric equations of the lines
For Line 1, we can express it in parametric form:
- Let \( t \) be the parameter for Line 1:
\[
x = 1 + 2t, \quad y = -1 + 3t, \quad z = 1 + 4t
\]
For Line 2, let \( s \) be the parameter:
- The parametric equations for Line 2 are:
\[
x = 3 + s, \quad y = k + 2s, \quad z = s
\]
### Step 3: Set the parametric equations equal to each other
To find the intersection, we set the equations of the two lines equal:
1. From \( x \):
\[
1 + 2t = 3 + s \quad \Rightarrow \quad s = 2t - 2 \quad \text{(Equation 1)}
\]
2. From \( y \):
\[
-1 + 3t = k + 2s \quad \Rightarrow \quad -1 + 3t = k + 2(2t - 2) \quad \Rightarrow \quad -1 + 3t = k + 4t - 4
\]
Rearranging gives:
\[
3t - 4t = k - 4 + 1 \quad \Rightarrow \quad -t = k - 3 \quad \Rightarrow \quad t = 3 - k \quad \text{(Equation 2)}
\]
3. From \( z \):
\[
1 + 4t = s \quad \Rightarrow \quad s = 1 + 4t \quad \text{(Equation 3)}
\]
### Step 4: Substitute and solve for \( k \)
Now we have three equations:
- From Equation 1: \( s = 2t - 2 \)
- From Equation 2: \( t = 3 - k \)
- From Equation 3: \( s = 1 + 4t \)
Substituting \( t = 3 - k \) into Equation 1:
\[
s = 2(3 - k) - 2 = 6 - 2k - 2 = 4 - 2k \quad \text{(Equation 4)}
\]
Substituting \( t = 3 - k \) into Equation 3:
\[
s = 1 + 4(3 - k) = 1 + 12 - 4k = 13 - 4k \quad \text{(Equation 5)}
\]
Now we set Equation 4 equal to Equation 5:
\[
4 - 2k = 13 - 4k
\]
Rearranging gives:
\[
4k - 2k = 13 - 4 \quad \Rightarrow \quad 2k = 9 \quad \Rightarrow \quad k = \frac{9}{2}
\]
### Step 5: Find the equation of the plane containing the lines
To find the equation of the plane containing the two lines, we can use the formula:
\[
\begin{vmatrix}
x - \alpha_1 & y - \beta_1 & z - \gamma_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{vmatrix} = 0
\]
Where \( (\alpha_1, \beta_1, \gamma_1) \) is a point on the first line, and \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) are direction ratios of the lines.
Using \( \alpha_1 = 1, \beta_1 = -1, \gamma_1 = 1 \) for Line 1 and direction ratios \( (2, 3, 4) \) for Line 1 and \( (1, 2, 1) \) for Line 2, we set up the determinant:
\[
\begin{vmatrix}
x - 1 & y + 1 & z - 1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{vmatrix} = 0
\]
### Step 6: Calculate the determinant
Calculating the determinant:
\[
(x - 1) \begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} - (y + 1) \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} + (z - 1) \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 0
\]
Calculating the minors:
1. \( \begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 - 4 \cdot 2 = 3 - 8 = -5 \)
2. \( \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 4 \cdot 1 = 2 - 4 = -2 \)
3. \( \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 2 \cdot 2 - 3 \cdot 1 = 4 - 3 = 1 \)
Substituting back gives:
\[
(x - 1)(-5) - (y + 1)(-2) + (z - 1)(1) = 0
\]
Expanding this:
\[
-5x + 5 + 2y + 2 + z - 1 = 0
\]
Combining like terms:
\[
-5x + 2y + z + 6 = 0 \quad \Rightarrow \quad 5x - 2y - z - 6 = 0
\]
### Final Answer
The value of \( k \) is \( \frac{9}{2} \) and the equation of the plane containing the lines is:
\[
5x - 2y - z - 6 = 0
\]
