Step 1 (Reason: Parallelism implies a common proportionality constant). Let \[ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k\neq 0. \]
Step 2 (Reason: rewrite \(a_1\) and \(c_1\) using \(k\)). From above, \(a_1=k\,a_2\) and \(c_1=k\,c_2\).
Step 3 (Reason: substitute and cancel). \[ \frac{a_1 c_2}{a_2}=\frac{k\,a_2\cdot c_2}{a_2}=k\,c_2=c_1. \] Hence \(\dfrac{a_1 c_2}{a_2}=c_1\).
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} $