For two lines to be coplanar, the scalar triple product of their direction ratios and the vector joining points on the lines must be zero. The direction ratios are \((1, 1, -k)\) and \((k, 2, 1)\). A point on the first line is \((2, 3, 4)\) and on the second is \((1, 4, 5)\). The vector joining them is \((-1, 1, 1)\). Setting up the determinant: \(\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0\). Expanding: \(-1(1+2k) - 1(1+k^2) + 1(2-k) = 0\), which simplifies to \(k^2 + 3k = 0\), giving \(k = 0\) or \(k = -3\). Thus, k has exactly two values.