1. Key Concepts and Formulas

• Condition for Intersecting Lines: Two lines in three-dimensional space intersect if and only if they are coplanar and not parallel. Mathematically, this means the shortest distance between them is zero. For two lines given in vector form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$, they intersect if and only if the scalar triple product of the vector connecting points on the lines and their direction vectors is zero.
• Scalar Triple Product (Determinant Form): If Line 1 passes through point $A_1(x_1, y_1, z_1)$ with direction ratios $(l_1, m_1, n_1)$ and Line 2 passes through point $A_2(x_2, y_2, z_2)$ with direction ratios $(l_2, m_2, n_2)$, then the condition for intersection is: $\begin{vmatrix} (x_2-x_1) & (y_2-y_1) & (z_2-z_1) \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
• Standard Form of a Line: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(l, m, n)$ is given by $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$.

2. Step-by-Step Solution

Step 1: Identify Points and Direction Ratios from the Given Lines

We are given two lines in Cartesian form. Our first step is to extract the coordinates of a point on each line and their respective direction ratios. This is crucial for setting up the determinant in the intersection condition.

• Line 1 ($L_1$): $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$

  • By comparing with the standard form $\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1}$, we identify:
    • A point on $L_1$: $A_1(x_1, y_1, z_1) = (1, 2, 3)$
    • Direction ratios of $L_1$: $(l_1, m_1, n_1) = (k, 2, 3)$

• Line 2 ($L_2$): $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$

  • Similarly, for $L_2$:
    • A point on $L_2$: $A_2(x_2, y_2, z_2) = (2, 3, 1)$
    • Direction ratios of $L_2$: $(l_2, m_2, n_2) = (3, k, 2)$

Step 2: Calculate the Components of the Vector Connecting the Points

Next, we need the components of the vector $\vec{A_1A_2}$, which is $(\vec{a_2} - \vec{a_1})$. These components form the first row of our determinant. This vector represents the displacement from a point on the first line to a point on the second line.

• The components are $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$.
  • $x_2 - x_1 = 2 - 1 = 1$
  • $y_2 - y_1 = 3 - 2 = 1$
  • $z_2 - z_1 = 1 - 3 = -2$
• So, the vector connecting the points is $\vec{A_1A_2} = \widehat{i} + \widehat{j} - 2\widehat{k}$.

Step 3: Apply the Intersection Condition using the Determinant

Now we use the condition for intersecting lines, which states that the determinant formed by the components of $\vec{A_1A_2}$ and the direction ratios of both lines must be zero. This is because if the lines intersect, these three vectors are coplanar.

• Substitute the values obtained in Step 1 and Step 2 into the determinant formula: $\begin{vmatrix} (x_2-x_1) & (y_2-y_1) & (z_2-z_1) \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ $\begin{vmatrix} 1 & 1 & -2 \\ k & 2 & 3 \\ 3 & k & 2 \end{vmatrix} = 0$

Step 4: Expand the Determinant and Form a Quadratic Equation

We expand the $3 \times 3$ determinant to obtain an algebraic equation in terms of $k$. Remember the expansion formula for a $3 \times 3$ determinant: $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)$

• Applying this to our determinant: $1 \cdot ((2)(2) - (3)(k)) - 1 \cdot ((k)(2) - (3)(3)) + (-2) \cdot ((k)(k) - (2)(3)) = 0$ $1(4 - 3k) - 1(2k - 9) - 2(k^2 - 6) = 0$

• Simplify the expression: $4 - 3k - 2k + 9 - 2k^2 + 12 = 0$

• Combine like terms to form a quadratic equation: $-2k^2 + (-3k - 2k) + (4 + 9 + 12) = 0$ $-2k^2 - 5k + 25 = 0$

• Multiply by -1 to make the leading coefficient positive: $2k^2 + 5k - 25 = 0$

Step 5: Solve the Quadratic Equation for $k$

We now solve the quadratic equation $2k^2 + 5k - 25 = 0$. We can use factoring or the quadratic formula. Factoring is often quicker if the factors are easily identifiable. We look for two numbers that multiply to $2 \times (-25) = -50$ and add up to $5$. These numbers are $10$ and $-5$.

• Rewrite the middle term using these numbers: $2k^2 + 10k - 5k - 25 = 0$

• Factor by grouping: $2k(k + 5) - 5(k + 5) = 0$ $(2k - 5)(k + 5) = 0$

• This gives two possible values for $k$:

  • $2k - 5 = 0 \Rightarrow 2k = 5 \Rightarrow k = \frac{5}{2}$
  • $k + 5 = 0 \Rightarrow k = -5$

Step 6: Select the Integer Value of $k$

The problem specifically asks for the integer value of $k$.

• Comparing our two solutions:
  • $k = \frac{5}{2}$ is not an integer.
  • $k = -5$ is an integer.

Therefore, the integer value of $k$ for which the lines intersect is $-5$.

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting coordinates and direction ratios, especially if the line equation is given in a non-standard form (e.g., $\frac{1-x}{l}$ instead of $\frac{x-1}{-l}$). Also, pay close attention to signs during determinant expansion and algebraic simplification.
• Algebraic Precision: Solving the quadratic equation accurately is crucial. Double-check all calculations, particularly when combining terms and factoring.
• Read the Question Carefully: Always note any specific constraints, like $k$ being an integer, as this helps in selecting the correct answer from multiple solutions.

4. Summary

To determine the integer value of $k$ for which the two given straight lines intersect, we employed the fundamental condition for intersecting lines in 3D geometry. This condition states that the scalar triple product of the vector connecting a point on each line and their respective direction vectors must be zero. We extracted the necessary points and direction ratios from the Cartesian equations of the lines, set up the $3 \times 3$ determinant, and expanded it to form a quadratic equation in $k$. Solving this quadratic equation yielded two possible values for $k$. Finally, by applying the problem's constraint that $k$ must be an integer, we selected the appropriate solution.

The final answer is $\boxed{-5}$, which corresponds to option (A).