If the line $\frac{-x+1}{3}=\frac{-y-2}{-2k}=\frac{z+3}{2}$ and $\frac{-1+x}{3k}=\frac{-1+ y}{1}=\frac{-z+6}{5}$ are perpendicular, then the value of $k$ is:

$-\frac{10}{7}$

The correct answer is Option (2) → $-\frac{10}{7}$

First line:

$-\frac{x+1}{3}=\frac{-y-2}{-2k}=\frac{z+3}{2}=t$

$x=-3t-1,\quad y=2kt-2,\quad z=2t-3$

Direction ratios: $(-3,\;2k,\;2)$

Second line:

$\frac{-1+x}{3k}=\frac{-1+y}{1}=\frac{-z+6}{5}=s$

$x=3ks+1,\quad y=s+1,\quad z=-5s+6$

Direction ratios: $(3k,\;1,\;-5)$

Condition for perpendicularity:

$(-3)(3k)+(2k)(1)+(2)(-5)=0$

$-9k+2k-10=0$

$-7k-10=0$

$k=-\frac{10}{7}$