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

-1

The correct answer is Option (2) → -1

first making std. form of lines

$l_1:\frac{x-3}{k-5}=\frac{y-1}{1}=\frac{5-z}{-2k-1}$

$l_2:\frac{x+2}{-1}=\frac{2-y}{-k}=\frac{z}{5}$

$l_1⊥l_2$

so $((k-5)\hat i+\hat j+(2k+1)\hat k).(-\hat i+k\hat j+5\hat k)=0$

$5-k+k+10k+5=0$

$k=-1$