The lines $\frac{1-x}{2} = \frac{y-1}{3} = \frac{z}{1}$ and $\frac{2x-3}{2p} = \frac{y}{-1} = \frac{z-4}{7}$ are perpendicular to each other for $p$ equal to : (A) $-\frac{1}{2}$ (B) $\frac{1}{2}$ (C) 2 (D) 3