Discussion
The lines \(\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}\) and \(\frac{x-4}{1} = \frac{y+3}{k} = \frac{z+1}{7}\) are coplanar. Then the value of k is
Answer: E
\(k = -4\).
Two lines through points \(A\) and \(B\) with directions \(\vec{b}_{1}, \vec{b}_{2}\) are coplanar when \(\overrightarrow{AB}\), \(\vec{b}_{1}\) and \(\vec{b}_{2}\) are coplanar, that is their scalar triple product is zero. Here \(A = (1, -1, -10)\), \(B = (4, -3, -1)\), so \(\overrightarrow{AB} = (3, -2, 9)\):
\[ \begin{aligned} \begin{vmatrix} 3 & -2 & 9 \\ 2 & -3 & 8 \\ 1 & k & 7 \end{vmatrix} &= 0 \\ 3(-21 - 8k) + 2(14 - 8) + 9(2k + 3) &= 0 \\ -24 - 6k &= 0, \quad k = -4 \end{aligned} \]
The positive value 4 is the sign slip in the last line; 0, 2 and \(-2\) leave the determinant nonzero, so those lines would be skew.
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