If the line $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-7}$ are perpendicular to each other, find the value of $k$.

$-2$

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

The equation of the given lines are

$L_1: \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$

$L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-7}$

Their Direction Ratio are $(-3, 2k, 2)$ and $(3k, 1, -7)$

when $L_1 \perp L_2$, $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

$(-3)(+3k) + (2k)(1) + (2)(-7) = 0$

$-9k + 2k - 14 = 0 \Rightarrow -7k - 14 = 0 \Rightarrow k = -2$