To solve this, recall the properties of determinants and adjugates. If \( A \) is a 3×3 matrix, then \( |\text{adj}(A)| = |A|^2 \).
Similarly, for the adjugate of a scalar multiple of a matrix, use the identity \( \text{adj}(kA) = k^{n-1} \text{adj}(A) \),
where \( n \) is the order of the matrix.
Solve for \( B \) and then compute the trace and determinant to find the solution.
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $