Step 1: Understanding the Concept:
First, observe the direction ratios of the two lines.
Line 1 has direction ratios $(2, 3, 4)$.
Line 2 has direction ratios $(4, 6, 8)$, which can be simplified to $(2, 3, 4)$ by dividing by 2.
Since their direction ratios are proportional, the two lines are parallel.
We need to use the formula for the shortest distance between two parallel lines.
Step 2: Key Formula or Approach:
The distance $d$ between two parallel lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}$ is given by:
\[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \]
Step 3: Detailed Explanation:
Let's extract the necessary vectors from the line equations.
Line 1 passes through $\vec{a}_1 = k\hat{i} + 4\hat{j} + 3\hat{k}$ and is parallel to $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Line 2 passes through $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 7\hat{k}$. It is also parallel to $\vec{b}$ since $(4,6,8)$ is parallel to $(2,3,4)$.
Now, calculate the vector difference $\vec{a}_2 - \vec{a}_1$:
\[ \vec{a}_2 - \vec{a}_1 = (2 - k)\hat{i} + (4 - 4)\hat{j} + (7 - 3)\hat{k} = (2 - k)\hat{i} + 0\hat{j} + 4\hat{k} \]
Next, find the cross product $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$:
\[ (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2-k & 0 & 4
2 & 3 & 4 \end{vmatrix} \]
\[ = \hat{i}(0 \cdot 4 - 4 \cdot 3) - \hat{j}((2-k) \cdot 4 - 4 \cdot 2) + \hat{k}((2-k) \cdot 3 - 0 \cdot 2) \]
\[ = \hat{i}(-12) - \hat{j}(8 - 4k - 8) + \hat{k}(6 - 3k) \]
\[ = -12\hat{i} + 4k\hat{j} + (6 - 3k)\hat{k} \]
Now find the magnitude of this cross product:
\[ |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-12)^2 + (4k)^2 + (6 - 3k)^2} \]
\[ = \sqrt{144 + 16k^2 + (36 - 36k + 9k^2)} \]
\[ = \sqrt{25k^2 - 36k + 180} \]
Find the magnitude of the direction vector $\vec{b}$:
\[ |\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]
We are given that the shortest distance $d = \frac{13}{\sqrt{29}}$.
\[ \frac{\sqrt{25k^2 - 36k + 180}}{\sqrt{29}} = \frac{13}{\sqrt{29}} \]
Equating the numerators and squaring:
\[ 25k^2 - 36k + 180 = 13^2 = 169 \]
\[ 25k^2 - 36k + 11 = 0 \]
Factorize the quadratic equation:
\[ 25k^2 - 25k - 11k + 11 = 0 \]
\[ 25k(k - 1) - 11(k - 1) = 0 \]
\[ (25k - 11)(k - 1) = 0 \]
This gives $k = 1$ or $k = \frac{11}{25}$.
Comparing with the given options, $k = 1$ is the correct choice.
Step 4: Final Answer:
The value of $k$ is $1$.