<p>The lines \(\dfrac{x-2}{1} = \dfrac{y-3}{1} = \dfrac{z-4}{-k}\) and \(\dfrac{x-1}{k} = \dfrac{y-4}{2} = \dfrac{z-5}{1}\) are coplanar is</p>
<p>\(k = 0\) or \(-1\)</p>
<p>\(k = 1\) or \(-1\)</p>
<p>\(k = 0\) or \(-3\)</p>
<p>\(k = 3\) or \(-3\)</p>
Step-by-Step Solution
Key Concept: Two lines are coplanar if and only if the scalar triple product of their direction vectors and the vector connecting any two points on them equals zero. This requires the determinant condition: the vector between points on the lines must be linearly dependent on both direction vectors.
Step 1: Identify the components. Line 1: Point A(2,3,4), direction vector d_1 = (1,1,-k). Line 2: Point B(1,4,5), direction vector d_2 = (k,2,1). Step 2: Find vector AB = B - A = (-1,1,1). Step 3: Apply coplanarity condition: AB · (d_1 × d_2) = 0. This means the scalar triple product determinant must equal zero: $\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0$ Step 4: Expand along the first row: -1(1·1 - (-k)·2) - 1(1·1 - (-k)·k) + 1(1·2 - 1·k) = 0 -1(1 + 2k) - 1(1 + k^2) + 1(2 - k) = 0 -1 - 2k - 1 - k^2 + 2 - k = 0 -k^2 - 3k = 0 -k(k + 3) = 0 Step 5: Therefore k = 0 or k = -3. The answer is k = 0 or k = -3 (typically option C presents both values or the non-trivial solution k = -3).
Correct Answer: C