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

$2$

The correct answer is Option (3) → $2$ ##

$L_1: \frac{1-x}{2} = \frac{y-1}{3} = \frac{z}{1} \Rightarrow \frac{x-1}{-2} = \frac{y-1}{3} = \frac{z}{1} \dots(i)$

$L_2: \frac{2x-3}{2p} = \frac{y}{-1} = \frac{z-4}{7} \Rightarrow \frac{x-\frac{3}{2}}{p} = \frac{y}{-1} = \frac{z-4}{7} \dots(ii)$

On Comparing with $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$

Direction ratio of line (i) are $a_1 = -2, b_1 = 3, c_1 = 1$

Direction ratio of line (ii) are $a_2 = p, b_2 = -1, c_2 = 7$

When $L_1 \perp L_2$, then $a_1a_2 + b_1b_2 + c_1c_2 = 0$

$-2 \times p + 3 \times (-1) + 1 \times 7 = 0$

$-2p - 3 + 7 = 0$

$-2p + 4 = 0$

$p = 2$