Key Concepts and Formulas

• Equation of a Line in Symmetric Form: A line passing through a point $P_1(x_1, y_1, z_1)$ and parallel to a direction vector $\vec{v_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ can be represented as: $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ From this form, we can directly identify a point on the line and its direction ratios.
• Condition for Coplanarity of Two Lines: Two lines $L_1$ and $L_2$ are coplanar if and only if the scalar triple product of the vector connecting a point on $L_1$ to a point on $L_2$, and their respective direction vectors, is zero. Given Line 1 ($L_1$): passes through $P_1(x_1, y_1, z_1)$ with direction vector $\vec{v_1} = (a_1, b_1, c_1)$. Given Line 2 ($L_2$): passes through $P_2(x_2, y_2, z_2)$ with direction vector $\vec{v_2} = (a_2, b_2, c_2)$. The condition for coplanarity is: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ Geometrically, this means the three vectors, $\vec{P_1P_2}$, $\vec{v_1}$, and $\vec{v_2}$, lie in the same plane.

Step-by-Step Solution

Step 1: Extract Points and Direction Vectors from the Given Lines We begin by identifying a point and the direction vector for each of the given lines by comparing them with the standard symmetric form $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$.

For the first line ($L_1$): ${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$

• The point $P_1(x_1, y_1, z_1)$ on this line is $(k, 2, 3)$.
• The direction vector $\vec{v_1} = (a_1, b_1, c_1)$ for this line is $(1, 2, 3)$.

For the second line ($L_2$): ${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$ To clearly identify the point coordinates, we rewrite the terms as $x - (-1)$, $y - (-2)$, and $z - (-3)$.

• The point $P_2(x_2, y_2, z_2)$ on this line is $(-1, -2, -3)$.
• The direction vector $\vec{v_2} = (a_2, b_2, c_2)$ for this line is $(3, 2, 1)$.

Step 2: Form the Vector Connecting the Points Next, we calculate the components of the vector $\vec{P_1P_2}$ (or $\vec{P_2P_1}$). We will use $(x_1-x_2, y_1-y_2, z_1-z_2)$ for the first row of our determinant.

• $x_1 - x_2 = k - (-1) = k + 1$
• $y_1 - y_2 = 2 - (-2) = 2 + 2 = 4$
• $z_1 - z_2 = 3 - (-3) = 3 + 3 = 6$ So, the vector connecting the points is $\vec{P_2P_1} = (k+1, 4, 6)$.

Step 3: Apply the Coplanarity Condition Now we substitute these values, along with the direction vector components, into the coplanarity determinant condition: $\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ $\begin{vmatrix} k+1 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = 0$ This equation is the algebraic translation of the geometric condition that the two lines are coplanar.

Step 4: Evaluate the Determinant We expand the $3 \times 3$ determinant along the first row: $(k+1) \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} - 4 \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = 0$ Calculate each $2 \times 2$ determinant:

• $(k+1) \times ((2 \times 1) - (3 \times 2)) = (k+1)(2 - 6) = (k+1)(-4)$
• $-4 \times ((1 \times 1) - (3 \times 3)) = -4(1 - 9) = -4(-8)$
• $+6 \times ((1 \times 2) - (2 \times 3)) = +6(2 - 6) = +6(-4)$

Substitute these results back into the equation: $(k+1)(-4) - 4(-8) + 6(-4) = 0$ $-4k - 4 + 32 - 24 = 0$

Step 5: Solve for k Simplify the equation and solve for $k$: $-4k + 4 = 0$ $-4k = -4$ $k = \frac{-4}{-4}$ $k = 1$

Common Mistakes & Tips

• Sign Errors in Point Extraction: Be extremely careful when identifying $x_1, y_1, z_1$ and $x_2, y_2, z_2$, especially with terms like $(x+1)$ which corresponds to $x-(-1)$.
• Determinant Calculation Errors: A small arithmetic mistake during the expansion of the $3 \times 3$ determinant can lead to an incorrect value of $k$. Double-check your calculations.
• Understanding the Condition: Remember that the determinant being zero means the three vectors ($\vec{P_1P_2}$, $\vec{v_1}$, $\vec{v_2}$) are linearly dependent, implying they lie in the same plane. This is the geometric basis for the coplanarity condition.

Summary

This problem effectively tests the understanding of the coplanarity condition for two lines in three-dimensional space. By first extracting the coordinates of a point on each line and their respective direction vectors, we form a determinant using these components. Setting this determinant equal to zero allows us to establish an algebraic equation involving the unknown parameter $k$. Solving this equation yields the value of $k$ that ensures the two lines lie on the same plane. The calculated value of $k=1$ confirms the coplanarity of the given lines.

The final answer is $\boxed{1}$.